A pair of dice is thrown 5 times. For each throw, a total of 5 is...

Given that,

A pair of dice is thrown $5$ times.

Required Sample space is \left\{\left(1, 1\right), \left(1, 2\right), \left(1, 3\right) . . . . . . . . . . . \left(6, 6\right)\right\}$</p><p>Total number of observations$= 36$</p><p>Now, a sum of$5$is observed in \left\{\left(1, 4\right), \left(4, 1\right), \left(2, 3\right), \left(3, 2\right)\right\}$

No of favourable outcomes $= 4$

$P$(success) $p = \dfrac{4}{36} = \dfrac{1}{9}$

$P$(failure) $q = 1 - \dfrac{1}{9} = \dfrac{8}{9}$

We know that for a binomial distribution, $B\left(n, p\right)$

$P\left(X = k\right) = Ck n {p}^{k}{q}^{n - k}$ where $k \ge 0$ and $p + q = 1$

$∴$Required probability

$⇒ P\left(X \ge 4\right) = P\left(X = 4\right) + P\left(X = 5\right)$

$= C4 5 {\left(\dfrac{1}{9}\right)}^{4}\dfrac{8}{9} + C5 5 {\left(\dfrac{1}{9}\right)}^{5}$

$= 5 \cdot \dfrac{8}{{9}^{5}} + \dfrac{1}{{9}^{5}} = \dfrac{41}{{9}^{5}}$

$= \dfrac{41 \times 3}{{3}^{10} \times 3} = \dfrac{123}{{3}^{11}}$

But given that $P\left(X \ge 4\right) = \dfrac{k}{{3}^{11}}$

$⇒ \dfrac{k}{{3}^{11}} = \dfrac{123}{{3}^{11}}$

$⇒ k = 123$

Hence, this is the correct option.