An integer is chosen form 2k /-9 ≤ k ≤ 10 . The probability that it is divisible by both 4 and 6 is

Question

An integer is chosen form 2k /-9 ≤ k ≤ 10 . The probability that it is divisible by both 4 and 6 is (A) (1/10) (B) (1/20) (C) (1/4) (D) (3/20)

Tardigrade Admin · Accepted Answer

Let S= 2 k mid-9 ≤ k ≤ 10 = -18,-16,-14, ldots, 0,2,4,6, ldots 20 Total number of possible outcomes =20 Favourable outcomes are -12,0 and 12 So, number of favourable outcomes =3 ∴ Required probability =(3/20)

Q. An integer is chosen form ${2 k / - 9 \leq k \leq 10}$ . The probability that it is divisible by both $4$ and $6$ is

$\frac{1}{10}$

$\frac{1}{20}$

$\frac{1}{4}$

$\frac{3}{20}$

Let $S = {2 k ∣ - 9 \leq k \leq 10} =$
${- 18, - 16, - 14, \dots, 0, 2, 4, 6, \dots 20}$
Total number of possible outcomes $= 20$
Favourable outcomes are $- 12, 0$ and 12
So, number of favourable outcomes $= 3$
$∴$ Required probability $= \frac{3}{20}$

Q. An integer is chosen form {2k/−9≤k≤10}. The probability that it is divisible by both 4 and 6 is

101​

201​

41​

203​