Question

$7$ persons are stopped on the road at random and asked about their birthdays. If the probability that $3$ of them are born on Wednesday, 2 on Thursday and the remaining $2$ on Sunday is $\frac{K}{7^6}$, then K is equal to

• A. $15$
• B. $30$
• C. $105$
• D. $210$

Answer 1

The correct option is B $30$

Given, 7 stopped at road and asked their birthdays.

Probability that 3 people out of 7 born on Wednesday $=$ Selecting 3 people among 7 divided by total ways$=$$\frac{{}^7{C}_3}{7^3}$

Probability that 2 people out of remaining 4, born on Thursday $=$selecting 2 people from remaining 4people=${}^4{C}_2{7}^2$

Probability of remaining 2 born on Sunday$=$${}^2{C}_2{7}^2$

$\therefore$ The required probability $=$ $\left(\frac{{}^7{C}_3}{7^3}\right)\ast \left({}^4{C}_2{7}^2\right)\ast \left({}^2{C}_2{7}^2\right)$

$\Rightarrow K{=}^7{C}_3{\ast }^4{C}_2{\ast }^2{C}_2$

$\Rightarrow 7K=35\ast 6\ast 1$

$\Rightarrow K=30$