The Mathematics of Maybe : Introduction to Probability — NCERT Solutions

There are 3 candies: strawberry, lemon, and mint.

Only 1 is strawberry.

So,

$$P(\text{strawberry})=\frac{1}{3}$$

The outfits are formed by choosing one shirt and one pair of pants.

So there are 6 possible outfits.

From the table of 1000 cases:

• Less than 4000 km: 20
• 4001 to 9000 km: 210
• 9001 to 14000 km: 325
• More than 14000 km: 445

The probability for each part is found by dividing the number of cases by 1000.

From the table, tyres lasting less than 4000 km occurred in 20 out of 1000 cases.

$$P(\text{less than 4000 km})=\frac{20}{1000}=0.02$$

Between 4000 and 14000 km means the cases in the groups 4001 to 9000 and 9001 to 14000.

Number of such cases:

$$210+325=535$$

So,

$$P=\frac{535}{1000}=0.535$$

More than 14000 km occurred in 445 out of 1000 cases.

So,

$$P(\text{more than 14000 km})=\frac{445}{1000}=0.445$$

The word PEACE has 5 letters: P, E, A, C, E.

For P, E or C:

• P occurs 1 time
• E occurs 2 times
• C occurs 1 time

Favourable outcomes = $1+2+1=4$
Total outcomes = 5

So,

$$P(\text{P, E or C})=\frac{4}{5}$$

This is the probability given by the chapter’s counting of letters. If the question is interpreted as “the letter is one of the distinct letters P, E, or C,” then the count is 4 out of 5 because E appears twice in the word.

In PEACE, there are 5 letters total and 2 E's.

Not E means P, A, or C.

Favourable outcomes = 3
Total outcomes = 5

So,

$$P(\text{not an E})=\frac{3}{5}$$

The arrow can land on any of the 8 numbers $1$ to $8$, all equally likely.

For 8, favourable outcomes = 1
Total outcomes = 8

So,

$$P(8)=\frac{1}{8}$$

Odd numbers from 1 to 8 are $1,3,5,7$.

• Favourable outcomes = 4
• Total outcomes = 8

So,

$$P(\text{odd number})=\frac{4}{8}=\frac{1}{2}$$

Numbers greater than 2 among 1 to 8 are $3,4,5,6,7,8$.

• Favourable outcomes = 6
• Total outcomes = 8

So,

$$P(\text{greater than 2})=\frac{6}{8}=\frac{3}{4}$$

All possible numbers are from 1 to 8, and every one of them is less than 9.

So the event is certain.

$$P(\text{number less than 9})=\frac{8}{8}=1$$

Multiples of 3 from 1 to 8 are $3$ and $6$.

• Favourable outcomes = 2
• Total outcomes = 8

So,

$$P(\text{multiple of 3})=\frac{2}{8}=\frac{1}{4}$$

The basket has 4 red balls and 5 blue balls, so total balls = 9.

There are 9 balls in total: 4 red, 3 green, and 2 blue.

To draw 2 blue balls without replacement:

• First blue: $\frac{2}{9}$
• Second blue: $\frac{1}{8}$

So,
$$P(\text{2 blue balls})=\frac{2}{9}\times\frac{1}{8}=\frac{2}{72}=\frac{1}{36}$$

So the probability is $\frac{1}{36}$.

An event with probability 0 is an impossible event. For example, when throwing a pair of 6-sided dice, an event such as getting a sum of 13 has probability 0, because the largest possible sum is 12.

An outcome with probability 1 is a certain outcome. For example, when a die is thrown, the outcome getting a number from 1 to 6 is certain, so its probability is 1.

For this experiment, the sample space is the set of all possible sums when two dice are thrown:
$$S=\{2,3,4,5,6,7,8,9,10,11,12\}$$

A simple event with probability 0 is impossible, such as getting a sum of 13.

A certain outcome with probability 1 is getting a number between 2 and 12 as the sum, since one of these must happen when two dice are thrown.

A prime number greater than 5, possible as a sum of two dice, can be 7 or 11.

Total possible outcomes when two dice are rolled = $6\times 6=36$.

Favourable outcomes for sum 7:
$(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ → 6 outcomes

Favourable outcomes for sum 11:
$(5,6),(6,5)$ → 2 outcomes

Total favourable outcomes = $6+2=8$

So,
$$P=\frac{8}{36}=\frac{2}{9}$$

The computed probability is $\frac{2}{9}$, which is the correct value.

Total balls = $4+3+2=9$.

We want the probability that the two balls are of different colours.

It is easier to use the complement:

• Probability that both balls are the same colour:
• both red: $\frac{4}{9}\times\frac{3}{8}=\frac{12}{72}$
• both green: $\frac{3}{9}\times\frac{2}{8}=\frac{6}{72}$
• both blue: $\frac{2}{9}\times\frac{1}{8}=\frac{2}{72}$

So,
$$P(\text{same colour})=\frac{12+6+2}{72}=\frac{20}{72}=\frac{5}{18}$$

Therefore,
$$P(\text{different colours})=1-\frac{5}{18}=\frac{13}{18}$$

The correct probability is $\frac{13}{18}$.

Three coins are tossed. The sample space has $2^3=8$ outcomes:
$$\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$$

We need outcomes where:

• the first coin is Heads, and
• there are exactly two heads in total.

These outcomes are:

• $HHT$
• $HTH$

So favourable outcomes = 2.

Thus,
$$P=\frac{2}{8}=\frac{1}{4}$$

The correct probability is $\frac{1}{4}$.

A four-digit number is formed using 1, 2, 3, 4 without repetition.

Total numbers possible = $4! = 24$.

For the number to be even, the last digit must be 2 or 4.

• If last digit is 2, the remaining 3 digits can be arranged in $3! = 6$ ways.
• If last digit is 4, the remaining 3 digits can be arranged in $3! = 6$ ways.

Favourable outcomes = $6+6=12$.

So,
$$P=\frac{12}{24}=\frac{1}{2}$$

The correct probability is $\frac{1}{2}$.

Each question has 4 options, with only 1 correct answer.

Probability of getting one question correct by guessing = $\frac{1}{4}$.

Probability of getting one question wrong = $\frac{3}{4}$.

We want exactly 2 correct answers out of 3.

Number of ways to choose which 2 questions are correct:
$$\binom{3}{2}=3$$

So,
$$P=\binom{3}{2}\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right) =3\cdot\frac{1}{16}\cdot\frac{3}{4} =\frac{9}{64}$$

The correct probability is $\frac{9}{64}$.

The box has balls numbered 1, 2, 3, 4.

For the experiment drawing a ball, recording it, then replacing it and drawing again, the sample space consists of ordered pairs:
$$S=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)\}$$

For the experiment drawing a ball, recording it, and then drawing again without replacement, the second draw cannot be the same ball as the first, so the sample space consists of ordered pairs with no repetition:
$$S=\{(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2),(3,4),(4,1),(4,2),(4,3)\}$$

These can be shown using tree diagrams with branches for each possible first draw and second draw.

With replacement, the two draws are independent. The sample space is:
$$\{11,12,13,14,21,22,23,24,31,32,33,34,41,42,43,44\}$$
where the first digit is the first draw and the second digit is the second draw.

Without replacement, the second draw cannot repeat the first ball. The sample space is:
$$\{12,13,14,21,23,24,31,32,34,41,42,43\}$$

These are all ordered pairs with different numbers.

For the experiment with replacement, there are $4$ choices for the first draw and $4$ choices for the second draw:
$$4\times 4=16$$

For the experiment without replacement, there are $4$ choices for the first draw and $3$ for the second draw:
$$4\times 3=12$$

So the sizes of the two sample spaces are 16 and 12.

The coin has outcomes H and T. The card can be any of 1, 2, 3, 4, 5, 6.

So the sample space is the set of all ordered pairs:
$$\{H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6\}$$

There are 12 elements in all.

Three coins can give 0, 1, 2, or 3 heads. So the sample space for the number of heads is
$$\{0,1,2,3\}$$

The other lists fail because:

• $\{1,2,3\}$ does not include 0 heads.
• $\{0,1,2\}$ does not include 3 heads.
• $\{0,1,2,3,4\}$ includes 4, which is impossible.

So the correct sample space is $\{0,1,2,3\}$.

$\{1,2,3\}$ is not a complete sample space for the number of heads when three coins are tossed, because it misses the outcome 0 heads.

$\{0,1,2\}$ is not a complete sample space for the number of heads when three coins are tossed, because it misses the outcome 3 heads.

$\{0,1,2,3,4\}$ is not a correct sample space because 4 heads is impossible when only three coins are tossed.

$\{0,1,2,3\}$ is the correct sample space for the number of heads when three coins are tossed, because all possible outcomes are included and no impossible outcome is added.

The question refers to Fig. 7.8, but the dimensions of the rectangular region are not provided in the chapter text here.

The chapter says equally likely outcomes occur in fair situations.

From the listed examples, these have equally likely outcomes:

• Tossing a fair coin once
• Rolling a fair 6-sided die

The others do not necessarily have equally likely outcomes because the result may depend on the condition of the car, the marble counts, or the baby’s sex is not treated as a fair random experiment in this chapter.

The table gives 1000 cases in total.

(i) Less than 4000 km:
$$P=\frac{20}{1000}=0.02$$

(ii) Between 4000 and 14000 km: this corresponds to the two groups 4001 to 9000 and 9001 to 14000.
$$P=\frac{210+325}{1000}=\frac{535}{1000}=0.535$$

(iii) More than 14000 km:
$$P=\frac{445}{1000}=0.445$$

So the probabilities are 0.02, 0.535, and 0.445.