A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If probability of at least 4 suc

Question

A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If probability of at least 4 successes is k311, then k is equal to:

Accepted Answer

Given: A success means getting a sum of $5$ on one throw of a pair of dice. The pair of dice is thrown $5$ times.

Find: The value of $k$ if the probability of at least $4$ successes is written as $\frac{k}{3^{11}}$ .

The possible outcomes for getting a sum of $5$ are $\{(1,4),(2,3),(3,2),(4,1)\}$ . Hence,

P(\text{success})=\frac{4}{36}=\frac{1}{9}

and

P(\text{failure})=1-\frac{1}{9}=\frac{8}{9}

Now, probability of at least $4$ successes is

P(\text{at least }4\text{ successes})=P(4\text{ successes})+P(5\text{ successes})

Using the binomial probability formula,

P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}

with $n=5$ and $p=\frac{1}{9}$ .

So,

P(4\text{ successes})=\binom{5}{4}\left(\frac{1}{9}\right)^4\left(\frac{8}{9}\right)=5\cdot \frac{1}{9^4}\cdot \frac{8}{9}=\frac{40}{9^5}

and

P(5\text{ successes})=\binom{5}{5}\left(\frac{1}{9}\right)^5=\frac{1}{9^5}

Therefore,

P(\text{at least }4\text{ successes})=\frac{40}{9^5}+\frac{1}{9^5}=\frac{41}{9^5}

Now,

9^5=(3^2)^5=3^{10}

so

\frac{41}{9^5}=\frac{41}{3^{10}}=\frac{41\times 3}{3^{11}}=\frac{123}{3^{11}}

Hence, $k=123$ . The correct option is B.

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