A box contains cards bearing numbers 6 to 70.

Question:

A box contains cards bearing numbers 6 to 70. If one card is drawn at random from the box, find the probability that it bears

(i) a one-digit number,
(ii) a number divisible by 5,
(iii) an odd number less than 30,
(iv) a composite number between 50 and 70.

Solution:

​Given number 6, 7, 8, .... , 70 form an AP with a = 6 and d = 1.
Let Tn = 70. Then,
6 + (n − 1)1 = 70
⇒ 6 + n  − 1 = 70
⇒ n = 65

Thus, total number of outcomes = 65.

(i) Let E1 be the event of getting a one-digit number.

Out of these numbers, one-digit numbers are 6, 7, 8 and 9.

Number of favourable outcomes = 4.

$\therefore P$ (getting a one-digit number) $=P\left(E_{1}\right)=\frac{\text { Number of outcomes favourable to } E_{1}}{\text { Number of all possible outcomes }}$

$=\frac{4}{65}$

Thus, the probability that the card bears a one-digit number is $\frac{4}{65}$.

(ii) Let E2 be the event of getting a number divisible by 5.

Out of these numbers, numbers divisible by 5 are 10, 15, 20, ... , 70.
Given number 10, 15, 20, .... , 70 form an AP with a = 10 and d = 5.
Let Tn = 70. Then,
10 + (n − 1)5 = 70
⇒ 10 + 5n  − 5 = 70
⇒ 5n = 65
⇒ n = 13

Thus, number of favourable outcomes = 13.

$\therefore \mathrm{P}$ (getting a number divisible by 5$)=\mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{\text { Number of outcomes favourable to } \mathrm{E}_{2}}{\text { Number of all possible outcomes }}$

$=\frac{13}{65}=\frac{1}{5}$

Thus, the probability that the card bears a number divisible by 5 is $\frac{1}{5}$.

(iii) Let E3 be the event of getting an odd number less than 30.

Out of these numbers, odd numbers less than 30 are 7, 9, 11, ... , 29.
Given number 7, 9, 11, .... , 29 form an AP with a = 7 and d = 2.
Let Tn = 29. Then,
7 + (n − 1)2 = 29
⇒ 7 + 2n  − 2 = 29
⇒ 2n = 24
⇒ n = 12

Thus, number of favourable outcomes = 12.

$\therefore P($ getting an odd number less than 30$)=P\left(E_{3}\right)=\frac{\text { Number of outcomes favourable to } E_{3}}{\text { Number of all possible outcomes }}$

$=\frac{12}{65}$

Thus, the probability that the card bears an odd number less than 30 is $\frac{12}{65}$.

(iv) Let E4 be the event of getting a composite number between 50 and 70.

Out of these numbers, composite numbers between 50 and 70 are 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68 and 69.

Number of favourable outcomes = 15.

$\therefore P($ getting a composite number between 50 and 70$)=P\left(E_{4}\right)=\frac{\text { Number of outcomes favourable to } E_{4}}{\text { Number of all possible outcomes }}$

$=\frac{15}{65}=\frac{3}{13}$

Thus, the probability that the card bears a composite number between 50 and 70 is $\frac{3}{13}$.

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