If the probability that a randomly chosen 6-digit number formed by using digits 1 and 8 only is a multiple of 21 is p, then 96 p is equal to .

Question

If the probability that a randomly chosen 6-digit number formed by using digits 1 and 8 only is a multiple of 21 is p, then 96 p is equal to . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

2 × 2 × 2 × 2 × 2 × 2=64 Divisible by 21 when divided by 3 . Case - I: All 1 arrow (1) Case - II: All 8 arrow (1) Case - III: 3 ones 3 eights (6 !/3 ! × 3 !)=20 Required probability ∴ p =(22/64) 96 p=96 × (22/64)=33

Q. If the probability that a randomly chosen $6$ -digit number formed by using digits $1$ and $8$ only is a multiple of $21$ is $p$ , then $96 p$ is equal to ___.

$2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$
Divisible by 21 when divided by $3$ .
Case $- I :$ All $1 \to (1)$
Case - II : All $8 \to (1)$
Case - III : 3 ones & 3 eights
$\frac{6 !}{3 ! \times 3 !} = 20$
Required probability $∴ p = \frac{22}{64}$
$96 p = 96 \times \frac{22}{64} = 33$

Q. If the probability that a randomly chosen 6-digit number formed by using digits 1 and 8 only is a multiple of 21 is p, then 96p is equal to ___.

2×2×2×2×2×2=64 Divisible by 21 when divided by 3 . Case −I: All 1→(1) Case - II : All 8→(1) Case - III : 3 ones & 3 eights 3!×3!6!​=20 Required probability ∴p=6422​ 96p=96×6422​=33

Q. If the probability that a randomly chosen $6$ -digit number formed by using digits $1$ and $8$ only is a multiple of $21$ is $p$ , then $96 p$ is equal to ___.

$2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$
Divisible by 21 when divided by $3$ .
Case $- I :$ All $1 \to (1)$
Case - II : All $8 \to (1)$
Case - III : 3 ones & 3 eights
$\frac{6 !}{3 ! \times 3 !} = 20$
Required probability $∴ p = \frac{22}{64}$
$96 p = 96 \times \frac{22}{64} = 33$