Find the probability of getting 5 exactly twice in 7 throws of a die.

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Find the probability of getting 5 exactly twice in 7 throws of a die.

Solutions:

Let’s say $X$ represents the number of times you’ve gotten 5 in 7 die throws. The Bernoulli trials are also the repeated tossing of a die.

Thus, probability of getting 5 in a single throw, $p=1 / 6$

And, $q=1-p$

$=1-1 / 6$

$=5 / 6$

Clearly, we have $\mathrm{X}$ has the binomial distribution where $\mathrm{n}=7$ and $\mathrm{p}=1 / 6$

$\therefore P(X=x)={ }^{n} C_{x} q^{n-x} p^{x}$

$={ }^{7} C_{x}\left(\frac{5}{6}\right)^{7-x} \cdot\left(\frac{1}{6}\right)^{x}$

As a result, the chances of getting 5 exactly twice in a die are slim. $=\mathrm{P}(\mathrm{X}=2)$

$={ }^{7} C_{2}\left(\frac{5}{6}\right)^{5} \cdot\left(\frac{1}{6}\right)^{2}$

$=21 \cdot\left(\frac{5}{6}\right)^{5} \cdot \frac{1}{36}$

$=\left(\frac{7}{12}\right)\left(\frac{5}{6}\right)^{5}$

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