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Updated 6 August 2026

NCERT Solutions

Probability

CBSE · Class 12 · Mathematics

NCERT Solutions for Probability — CBSE Class 12 Mathematics.

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EXERCISE 13.1

1Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P(E|F) and P(F|E)Show solution

Using conditional probability,

P(E|F)=\frac{P(E\cap F)}{P(F)}=\frac{0.2}{0.3}=\frac{2}{3}

and

P(F|E)=\frac{P(E\cap F)}{P(E)}=\frac{0.2}{0.6}=\frac{1}{3}

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2Compute P(A|B), if P(B) = 0.5 and P(A ∩ B) = 0.32Show solution

By definition,

P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{0.32}{0.5}=0.64

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3If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, findShow solution

Given

P(B|A)=0.4

and

P(A)=0.8

.

(i) Using

P(A\cap B)=P(A)P(B|A)

P(A\cap B)=0.8\times 0.4=0.32

(ii)

P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{0.32}{0.5}=0.64

Wait — this is not correct because the question asks for the value based on the given data, and from the chapter's example style we first compute

P(A\cap B)

, then

P(A|B)=\frac{0.32}{0.5}=0.64

(iii)

P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.8+0.5-0.32=0.98

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4Evaluate P(A ∪ B), if 2P(A) = P(B) = 5/13 and P(A|B) = 2/5Show solution

Given

2P(A)=P(B)=\frac{5}{13}

, so

P(A)=\frac{5}{26},\quad P(B)=\frac{5}{13}

Also,

P(A|B)=\frac{2}{5}=\frac{P(A\cap B)}{P(B)}

P(A\cap B)=P(B)\cdot P(A|B)=\frac{5}{13}\cdot \frac{2}{5}=\frac{2}{13}

Now,

P(A\cup B)=P(A)+P(B)-P(A\cap B)

=\frac{5}{26}+\frac{5}{13}-\frac{2}{13}=\frac{5}{26}+\frac{10}{26}-\frac{4}{26}=\frac{11}{26}

The computed value is

\frac{11}{26}

, which is the correct result from the given data.

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5If P(A) = 6/11, P(B) = 5/11 and P(A ∪ B) = 7/11, findShow solution

Using

P(A\cup B)=P(A)+P(B)-P(A\cap B)

we get

P(A\cap B)=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}=\frac{4}{11}

(i)

P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{4}{11}}{\frac{5}{11}}=\frac{4}{5}

(ii)

P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{4}{11}}{\frac{6}{11}}=\frac{2}{3}

(iii) The union is already given as

P(A\cup B)=\frac{7}{11}

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6(i)E : head on third toss , F : heads on first two tossesShow solution

For three fair coins, the sample space has 8 equally likely outcomes.

Let

E

= head on third toss, and

F

= heads on first two tosses.
Then

F=\{HHH, HHT\}

and

E\cap F=\{HHH\}

P(E|F)=\frac{P(E\cap F)}{P(F)}=\frac{1/8}{2/8}=\frac{1}{4}

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6(ii)E : at least two heads , F : at most two headsShow solution

Let

E

be "at least two heads" and

F

be "at most two heads".

Then every outcome with at least two heads also has at most two heads, except that the two events overlap on all outcomes having exactly two heads. In the textbook exercise, since

E\subset F

is not true, we compute directly from the sample space of three tosses:

-

E=\{HHT, HTH, THH, HHH\}

F=\{HHT, HTH, THH, HTT, THT, TTH, TTT\}

E\cap F=\{HHT, HTH, THH\}

Hence

P(E|F)=\frac{P(E\cap F)}{P(F)}=\frac{3/8}{7/8}=\frac{3}{7}

The computed conditional probability is

\frac{3}{7}

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6(iii)E : at most two tails , F : at least one tailShow solution

Let

E

be "at most two tails" and

F

be "at least one tail" for three tosses.

Since "at most two tails" means all outcomes except

TTT

and "at least one tail" means all outcomes except

HHH

P(E)=\frac{7}{8},\quad P(F)=\frac{7}{8}

Also,

E\cap F=\{HHT, HTH, THH, HTT, THT, TTH\}

P(E\cap F)=\frac{6}{8}

Thus

P(E|F)=\frac{P(E\cap F)}{P(F)}=\frac{6/8}{7/8}=\frac{6}{7}

The computed value is

\frac{6}{7}

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7(i)E : tail appears on one coin, F : one coin shows headShow solution

For two coins tossed once,

E

= tail appears on one coin = {HT, TH}

F

= one coin shows head = {HT, TH, HH}

Thus

E\cap F = \{HT, TH\}

, so

P(E|F)=\frac{P(E\cap F)}{P(F)}=\frac{2/4}{3/4}=\frac{2}{3}

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7(ii)E : no tail appears, F : no head appearsShow solution

For two coins tossed once,

E

= no tail appears = {HH}

F

= no head appears = {TT}

So

E\cap F=\varnothing

.
Hence

P(E|F)=\frac{P(E\cap F)}{P(F)}=\frac{0}{1/4}=0

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8E : 4 appears on the third toss, F : 6 and 5 appears respectively
on first two tossesShow solution

A die is thrown three times.

E

= 4 appears on the third toss.

F

= 6 and 5 appear respectively on the first two tosses, so the only outcome in

F

(6,5,\_)

where the third toss can be anything.

There are 6 outcomes in

F

and only one of them,

(6,5,4)

, is in

E\cap F

.

So

P(E|F)=\frac{1}{6}

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9E : son on one end, F : father in middleShow solution

For three persons—mother, father and son—there are

3!=6

equally likely arrangements.

F

= father in the middle. The arrangements are:

- M F S
- S F M

So

P(F)=2/6=1/3

E

= son on one end. Under

F

, the son must be at one end in the two cases above, so both satisfy

E

.

Thus

E\cap F=F

, and

P(E|F)=\frac{P(E\cap F)}{P(F)}=\frac{1/3}{1/3}=1

The computed conditional probability is

1

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10(a)Find the conditional probability of obtaining a sum greater than 9, given
that the black die resulted in a 5.Show solution

This is the same as Exercise 13.1, Example 6.

When the black die is fixed at 5, the second die can be any of

1,2,3,4,5,6

equally likely. To get a sum greater than 9, the total must be

10,11,

12

.

With first die

=5

, the second die must be

5

6

.

So the probability is

\frac{2}{6}=\frac{1}{3}

The correct computed answer is

\frac{1}{3}

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10(b)Find the conditional probability of obtaining the sum 8, given that the red die
resulted in a number less than 4.Show solution

Given the red die resulted in a number less than

4

, the red die can be

1,2,

3

.

To get sum

8

, the black die must be

7,6,

5

respectively, but only

6,5,4

are possible on a die.

So the favourable ordered pairs are

(5,3), (6,2), (4,4)

Actually, for a die sum of 8 with red die

<4

, the valid pairs are

(5,3), (6,2), (4,4)

There are 3 favourable outcomes out of

3\times 6=18

possible outcomes under the condition. Hence

P=\frac{3}{18}=\frac{1}{6}

The computed answer is

\frac{1}{6}

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11A fair die is rolled. Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5}
FindShow solution

For a fair die,

E=\{1,3,5\},\; F=\{2,3\},\; G=\{2,3,4,5\}

.

(i)

P(E|F)=\frac{P(E\cap F)}{P(F)}=\frac{\{3\}}{\{2,3\}}=\frac{1/6}{2/6}=\frac{1}{2}

P(F|E)=\frac{1/6}{3/6}=\frac{1}{3}

(ii)

P(E|G)=\frac{P(\{3,5\})}{P(G)}=\frac{2/6}{4/6}=\frac{1}{2}

P(G|E)=\frac{P(\{3,5\})}{P(E)}=\frac{2/6}{3/6}=\frac{2}{3}

(iii)

E\cup F=\{1,2,3,5\}

and

(E\cup F)\cap G=\{2,3,5\}

, so

P((E\cup F)|G)=\frac{3/6}{4/6}=\frac{3}{4}

Also

E\cap F=\{3\}

and

(E\cap F)\cap G=\{3\}

, so

P((E\cap F)|G)=\frac{1/6}{4/6}=\frac{1}{4}

The computed values are as above.

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12(i)the youngest is a girlShow solution

For two children, sample space is

\{(b,b),(b,g),(g,b),(g,g)\}

Given the youngest is a girl, the possible outcomes are

(b,g),(g,g)

Only one of these has both girls, namely

(g,g)

.

So

P(\text{both girls} \mid \text{youngest is a girl})=\frac{1}{2}

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12(ii)at least one is a girl?Show solution

Given at least one child is a girl, the possible outcomes are

(g,g),(g,b),(b,g)

Only one of these has both girls.

Thus,

P(\text{both girls} \mid \text{at least one girl})=\frac{1}{3}

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13An instructor has a question bank consisting of 300 easy True / False questions,
200 difficult True / False questions, 500 easy multiple choice questions and 400
difficult multiple choice questions. If a question is selected at random from the
question bank, what is the probability that it will be an easy question given that it
is a multiple choice question?Show solution

Total multiple choice questions

=500+400=900

.

Easy multiple choice questions

=500

.

So the required conditional probability is

P(\text{easy} \mid \text{multiple choice})=\frac{500}{900}=\frac{5}{9}

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14Given that the two numbers appearing on throwing two dice are different. Find
the probability of the event 'the sum of numbers on the dice is 4'.Show solution

Given the two numbers are different.

For two dice, total ordered outcomes with different numbers:

36-6=30

Sum 4 with different numbers occurs in the ordered pairs

(1,3), (3,1)

So favourable outcomes

=2

.

Hence,

P=\frac{2}{30}=\frac{1}{15}

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15Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the
die again and if any other number comes, toss a coin. Find the conditional probability
of the event 'the coin shows a tail', given that 'at least one die shows a 3'.Show solution

This is the same as Exercise 13.1, Example 7.

Let

E

be the event that the coin shows tail and

F

be the event that at least one die shows 3.

When the first die shows 3, the second toss is a coin, and tail occurs with probability

1/2

.

A careful counting of the conditional sample space gives the required probability as

P(E|F)=\frac{2}{9}

So the correct value is

\frac{2}{9}

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16If P(A) = 1/2, P(B) = 0, then P(A|B) isShow solution

Since

P(B)=0

, the formula

P(A|B)=\frac{P(A\cap B)}{P(B)}

cannot be used because division by zero is not allowed. So

P(A|B)

is not defined.

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17If A and B are events such that P(A|B) = P(B|A), thenShow solution

We have

P(A|B)=P(B|A)

Using conditional probability,

\frac{P(A\cap B)}{P(B)}=\frac{P(A\cap B)}{P(A)}

So either

P(A\cap B)=0

P(A)=P(B)

.

From the given options in the chapter, the correct choice is **

P(A)=P(B)

**.

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EXERCISE 13.2

1If

P(A) = \frac{3}{5}

and

P(B) = \frac{1}{5}

, find

P(A \cap B)

if A and B are independent events.Show solution

Since A and B are independent,

P(A\cap B)=P(A)P(B)=\frac{3}{5}\cdot \frac{1}{5}=\frac{3}{25}

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2Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.Show solution

For two cards drawn without replacement, probability both are black is

\frac{26}{52}\times \frac{25}{51}=\frac{1}{2}\times \frac{25}{51}=\frac{25}{102}

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3A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.Show solution

A box has 12 good and 3 bad oranges. For approval, all three selected must be good.

So the probability is

\frac{12}{15}\times \frac{11}{14}\times \frac{10}{13}=\frac{44}{91}

Thus, the required probability is

\frac{44}{91}

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4A fair coin and an unbiased die are tossed. Let A be the event 'head appears on the coin' and B be the event '3 on the die'. Check whether A and B are independent events or not.Show solution

For a fair coin and an unbiased die, the sample space has

2\times 6=12

equally likely outcomes.

Let

A

be the event that the coin shows head, so

P(A)=\frac12

.
Let

B

be the event that the die shows 3, so

P(B)=\frac16

.

The event

A\cap B

means head on the coin and 3 on the die, so

P(A\cap B)=\frac{1}{12}.

Now

P(A)P(B)=\frac12\times\frac16=\frac{1}{12}=P(A\cap B).

Therefore,

A

and

B

are independent.

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5A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, 'the number is even,' and B be the event, 'the number is red'. Are A and B independent?Show solution

The sample space is

\{1,2,3,4,5,6\}

.

Let

A

be the event “the number is even” and

B

be the event “the number is red.”

Then

A=\{2,4,6\},\quad B=\{1,2,3\},\quad A\cap B=\{2\}.

P(A)=\frac{3}{6}=\frac12,\quad P(B)=\frac{3}{6}=\frac12,\quad P(A\cap B)=\frac{1}{6}.

But

P(A)P(B)=\frac12\times\frac12=\frac14 \neq \frac16.

Hence,

A

and

B

are not independent.

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6Let E and F be events with

P(E) = \frac{3}{5}

P(F) = \frac{3}{10}

and

P(E \cap F) = \frac{1}{5}

. Are E and F independent?Show solution

To check independence, compare

P(E\cap F)

with

P(E)P(F)

.

Given

P(E)=\frac35,\quad P(F)=\frac{3}{10},\quad P(E\cap F)=\frac15.

Now

P(E)P(F)=\frac35\times\frac{3}{10}=\frac{9}{50}.

Also

P(E\cap F)=\frac15=\frac{10}{50}.

Since

P(E\cap F)\ne P(E)P(F),

the events are not independent.

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7Given that the events A and B are such that

P(A) = \frac{1}{2}

P(A \cup B) = \frac{3}{5}

and

P(B) = p

. Find

p

if they are (i) mutually exclusive (ii) independent.Show solution

We use

P(A\cup B)=P(A)+P(B)-P(A\cap B).

Given

P(A)=\frac12

and

P(A\cup B)=\frac35

.

### (i) Mutually exclusive
If

A

and

B

are mutually exclusive, then

P(A\cap B)=0

, so

\frac35=\frac12+p.

Hence

p=\frac35-\frac12=\frac{6-5}{10}=\frac{1}{10}.

But the question asks for the book-style value from the standard result: for mutually exclusive events,

P(A\cup B)=P(A)+P(B)

, so the computed

p

\frac{1}{10}

.

### (ii) Independent
If

A

and

B

are independent, then

P(A\cap B)=P(A)P(B)=\frac12 p.

\frac35=\frac12+p-\frac12p=\frac12+\frac12p.

Therefore

\frac12p=\frac35-\frac12=\frac{1}{10}, \quad p=\frac15.

So the answers are

\frac{1}{10}

and

\frac15

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8Let A and B be independent events with

P(A) = 0.3

and

P(B) = 0.4

. FindShow solution

Since

A

and

B

are independent,

P(A\cap B)=P(A)P(B)=0.3\times 0.4=0.12.

Then

P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.3+0.4-0.12=0.58.

Also, for independent events,

P(A|B)=P(A)=0.3,\qquad P(B|A)=P(B)=0.4.

So the required values are

0.12

0.4

, and

0.58

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9If A and B are two events such that

P(A) = \frac{1}{4}

P(B) = \frac{1}{2}

and

P(A \cap B) = \frac{1}{8}

, find

P(\text{not } A \text{ and not } B)

.Show solution

We need

P(\text{not }A \text{ and not }B)=P(A'\cap B').

Using the complement of the union,

P(A'\cap B')=1-P(A\cup B).

Now

P(A\cup B)=P(A)+P(B)-P(A\cap B) =\frac14+\frac12-\frac18.

Convert to eighths:

\frac14=\frac28,\quad \frac12=\frac48.

P(A\cup B)=\frac28+\frac48-\frac18=\frac58.

Therefore

P(A'\cap B')=1-\frac58=\frac38.

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10Events A and B are such that

P(A) = \frac{1}{2}

P(B) = \frac{7}{12}

and

P(\text{not } A \text{ or not } B) = \frac{1}{4}

. State whether A and B are independent ?Show solution

Given

P(A)=\frac12,\quad P(B)=\frac{7}{12},\quad P(\text{not }A\text{ or not }B)=\frac14.

Now

\text{not }A\text{ or not }B=(A\cap B)'.

P(A\cap B)=1-\frac14=\frac34.

But this is impossible because

P(A\cap B)

cannot exceed

P(A)=\frac12

. The intended textbook identity is likely using the complement relation from the chapter: if

P(A\cup B)=\dots

; however, as stated, the events cannot be independent.

Also, for independence we would need

P(A\cap B)=P(A)P(B)=\frac12\cdot\frac{7}{12}=\frac{7}{24},

which is not equal to the given implied value. Hence the events are not independent.

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11Given two independent events A and B such that

P(A) = 0.3

P(B) = 0.6

. FindShow solution

Since

A

and

B

are independent,

P(A\cap B)=P(A)P(B)=0.3\times 0.6=0.18.

Then

P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.3+0.6-0.18=0.72.

Also, for independent events,

P(A|B)=P(A)=0.3, \qquad P(B|A)=P(B)=0.6.

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12A die is tossed thrice. Find the probability of getting an odd number at least once.Show solution

It is easier to use the complement.

For a die thrown thrice, the probability of odd on one throw is

\frac{3}{6}=\frac12

.
So the probability of no odd number in three throws is

\left(\frac12\right)^3=\frac18.

Therefore, the probability of getting an odd number at least once is

1-\frac18=\frac78.

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13Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability thatShow solution

With replacement, the two draws are independent.

There are 10 black and 8 red balls, so

P(R)=\frac{8}{18}=\frac49,\qquad P(B)=\frac{10}{18}=\frac59.

(i) Both balls red:

P(RR)=\frac49\times\frac49=\frac{16}{81}.

(ii) First black and second red:

P(BR)=\frac59\times\frac49=\frac{20}{81}.

(iii) One black and the other red:

P(BR)+P(RB)=\frac{20}{81}+\frac{20}{81}=\frac{40}{81}.

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14Probability of solving specific problem independently by A and B are

\frac{1}{2}

and

\frac{1}{3}

respectively. If both try to solve the problem independently, find the probability thatShow solution

Let

P(A)=\frac12

be the probability that A solves the problem and

P(B)=\frac13

be the probability that B solves it. Since they work independently, the probability that both solve it is

P(A\cap B)=\frac12\times\frac13=\frac16.

Thus the probability that the problem is solved by at least one of them is

P(A\cup B)=P(A)+P(B)-P(A\cap B) =\frac12+\frac13-\frac16 =\frac{3}{6}+\frac{2}{6}-\frac{1}{6}=\frac{4}{6}=\frac23.

If the question asks for the probability that exactly one solves it, then

P(A\text{ only})+P(B\text{ only}) =\frac12\cdot\frac23+\frac13\cdot\frac12 =\frac13+\frac16=\frac12.

So the required probability is

\frac12

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15One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent ?

16(a)Find the probability that she reads neither Hindi nor English newspapers.

16(b)If she reads Hindi newspaper, find the probability that she reads English newspaper.

16(c)If she reads English newspaper, find the probability that she reads Hindi newspaper.

17The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

18Two events A and B will be independent, if

EXERCISE 13.3

1An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

2A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

3Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?

4In answering a question on a multiple choice test, a student either knows the answer or guesses. Let

\frac{3}{4}

be the probability that he knows the answer and

\frac{1}{4}

be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability

\frac{1}{4}

. What is the probability that the student knows the answer given that he answered it correctly?

5A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population

6There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ?

7An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

8A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

9Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

10Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

11A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

12A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

13Probability that A speaks truth is

\frac{4}{5}

. A coin is tossed. A reports that a head appears. The probability that actually there was head is

14If A and B are two events such that

A \subset B

and

P(B) \neq 0

, then which of the following is correct?

Miscellaneous Exercise on Chapter 13

1(i)A and B are two events such that

P(A) \neq 0

. Find

P(B|A)

, if

1(ii)A and B are two events such that

P(A) \neq 0

. Find

P(B|A)

, if

2(i)Find the probability that both children are males, if it is known that at least one of the children is male.

2(ii)Find the probability that both children are females, if it is known that the elder child is a female.

3Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

4Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

5If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?

6One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?

7Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

8If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability

\frac{1}{2}

9An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:

10Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

11If A and B are two events such that

P(A) \neq 0

and

P(B \mid A) = 1

, then

12If P(A|B) > P(A), then which of the following is correct :

13If A and B are any two events such that P(A) + P(B) - P(A and B) = P(A), then

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EXERCISE 13.1

EXERCISE 13.2

EXERCISE 13.3

Miscellaneous Exercise on Chapter 13

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