Question

Seven chits are numbered $1$ to $7$. Four chits are drawn one by one with replacement. The probability that the least number appearing on any selected chit is $5$ , is

• A. ${\left(\frac{3}{7}\right)}^4$
• B. ${\left(\frac{6}{7}\right)}^3$
• C. $\frac{(5\times 4\times 3)}{7^3}$
• D. ${\left(\frac{3}{7}\right)}^3$

Answer 1

The correct option is A

${\left(\frac{3}{7}\right)}^4$

Explanation for the correct option.

Finding the probability that the least number appearing on any selected chit is $5$

Given: Seven chits are numbered$1$ to $7$ Four chits are drawn one by one with replacement.

Total number of chits $=7$

The event that the least number on any selected chit is $5$ [Given] is $\left\{5,6,7\right\}$

So, the probability of $5$ or $6$ or $7$ in one draw $=\frac{3}{7}$

Therefore, in each of $4$ draws, the probability that the chit contains $5,6,7$ is ${\left(\frac{3}{7}\right)}^4.$

Hence the correct answer is option (D)