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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 a number divisible by 5.
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.
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.
∴ P(getting a number divisible by 5) = P(E2) = `("Number of outcomes favourable to" E_2)/"Number of all possible outcomes"`
`= 13/65 = 1/5`
Thus, the probability that the card bears a number divisible by 5 is `1/5`.
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