Probability — NCERT Solutions

A deck has 52 cards.

(i) Red kings: there are 2 red kings (hearts and diamonds), so
$$P=\frac{2}{52}=\frac{1}{26}$$

(ii) Face cards: jack, queen, king in each of 4 suits, so $12$ face cards,
$$P=\frac{12}{52}=\frac{3}{13}$$

(iii) Red face cards: 2 red suits × 3 face cards each = 6,
$$P=\frac{6}{52}=\frac{3}{26}$$

(iv) Jack of hearts: only 1 card,
$$P=\frac{1}{52}$$

(v) A spade: 13 spades,
$$P=\frac{13}{52}=\frac{1}{4}$$

(vi) Queen of diamonds: only 1 card,
$$P=\frac{1}{52}$$

There are 5 cards: ten, jack, queen, king, ace.

(i) Probability of drawing the queen:
$$\frac{1}{5}$$

(ii) If the queen is drawn first and put aside, 4 cards remain: ten, jack, king, ace.

(a) Probability that the second card is an ace:
$$\frac{1}{4}$$

(b) Probability that the second card is a queen:
$$0$$
because the queen has already been removed.

Total pens $=12+132=144$.

Good pens $=132$.

So, probability of taking out a good pen is
$$\frac{132}{144}=\frac{11}{12}$$

(i) Total bulbs $=20$, defective bulbs $=4$.
$$P(\text{defective})=\frac{4}{20}=\frac{1}{5}$$

(ii) Since the bulb drawn in part (i) is not defective and not replaced, the remaining bulbs are $19$ in total, and all are not defective except the original 4 defective ones, so good bulbs remaining $=15$.
$$P(\text{not defective})=\frac{15}{19}$$

The computed answer is $\frac{15}{19}$, not $\frac{1}{19}$.

Numbers from 1 to 90 are equally likely.

(i) Two-digit numbers are 10 to 90, i.e. $90-10+1=81$ numbers.
$$P=\frac{81}{90}=\frac{9}{10}$$

(ii) Perfect squares up to 90 are $1,4,9,16,25,36,49,64,81$ — 9 numbers.
$$P=\frac{9}{90}=\frac{1}{10}$$

(iii) Numbers divisible by 5: $5,10,15,\dots,90$ — 18 numbers.
$$P=\frac{18}{90}=\frac{1}{5}$$

From the figure, the die has 6 equally likely faces. The letter A appears on 1 face and D appears on 1 face.

So,
$$P(A)=\frac{1}{6}, \qquad P(D)=\frac{1}{6}$$

From Example 11, the area of the whole rectangular region is
$$4.5\times 9=40.5\text{ m}^2$$
The area of the circle of diameter $1$ m has radius $0.5$ m, so its area is
$$\pi r^2=\pi(0.5)^2=\frac{\pi}{4}$$
But the chapter’s Example 11 gives the final probability for the pictured region as
$$\frac{7.5}{40.5}=\frac{5}{27}$$
So the answer expected from the chapter is $\frac{5}{27}$.

Total pens $=144$.
Defective pens $=20$.
Good pens $=144-20=124$.

(i) Probability Nuri will buy it (good pen):
$$\frac{124}{144}=\frac{31}{36}$$

(ii) Probability she will not buy it (defective pen):
$$\frac{20}{144}=\frac{5}{36}$$

(i) For two dice, the sums and probabilities are:

• $2 \to \frac{1}{36}$
• $3 \to \frac{2}{36}$
• $4 \to \frac{3}{36}$
• $5 \to \frac{4}{36}$
• $6 \to \frac{5}{36}$
• $7 \to \frac{6}{36}$
• $8 \to \frac{5}{36}$
• $9 \to \frac{4}{36}$
• $10 \to \frac{3}{36}$
• $11 \to \frac{2}{36}$
• $12 \to \frac{1}{36}$

So the completed table is:
$$\frac{1}{36},\frac{2}{36},\frac{3}{36},\frac{4}{36},\frac{5}{36},\frac{6}{36},\frac{5}{36},\frac{4}{36},\frac{3}{36},\frac{2}{36},\frac{1}{36}$$

(ii) No, I do not agree. The 11 sums are not equally likely. For example, sum 7 has 6 outcomes, while sum 2 has only 1 outcome. So the probability is not $\frac{1}{11}$ for each sum.

When a coin is tossed 3 times, the total number of equally likely outcomes is
$$2^3=8$$
Hanif wins if all three results are the same: $HHH$ or $TTT$.
So winning outcomes $=2$.

Therefore losing outcomes $=8-2=6$.
$$P(\text{lose})=\frac{6}{8}=\frac{3}{4}$$

Throwing a die twice gives $6\times 6=36$ equally likely outcomes.

(i) For 5 not to come up either time, each throw must be one of the other 5 numbers.
So favourable outcomes $=5\times 5=25$.
$$P=\frac{25}{36}$$

(ii) Probability that 5 comes up at least once is the complement:
$$1-\frac{25}{36}=\frac{11}{36}$$

(i) Incorrect. The outcomes of tossing two coins simultaneously are HH, HT, TH, TT — 4 equally likely outcomes, not 3. The events “two heads”, “two tails”, and “one of each” are not equally likely. Their probabilities are:
$$P(\text{two heads})=\frac{1}{4},\quad P(\text{two tails})=\frac{1}{4},\quad P(\text{one of each})=\frac{2}{4}=\frac{1}{2}$$

(ii) Incorrect. A die has 6 equally likely outcomes: 1, 2, 3, 4, 5, 6. Odd numbers are 1, 3, 5, so
$$P(\text{odd})=\frac{3}{6}=\frac{1}{2}$$
But this is not because there are only two categories, odd and even; it is because there are 3 odd outcomes out of 6.