Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Question:

Find the probability of throwing at most 2 sixes in 6 throws of a single die.

Solution:

The repeated tossing of the die are Bernoulli trials. Let X represent the number of times of getting sixes in 6 throws of the die.

Probability of getting six in a single throw of die, $p=\frac{1}{6}$

$\therefore q=1-p=1-\frac{1}{6}=\frac{5}{6}$

Clearly, X has a binomial distribution with n = 6

$\therefore \mathrm{P}(\mathrm{X}=x)={ }^{1} \mathrm{C}_{x} q^{n-x} p^{x}={ }^{6} \mathrm{C}_{x}\left(\frac{5}{6}\right)^{6-x} \cdot\left(\frac{1}{6}\right)^{x}$

P (at most 2 sixes) = P(X ≤ 2)

$=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)$

$={ }^{6} \mathrm{C}_{0}\left(\frac{5}{6}\right)^{6}+{ }^{6} \mathrm{C}_{1} \cdot\left(\frac{5}{6}\right)^{5} \cdot\left(\frac{1}{6}\right)+{ }^{6} \mathrm{C}_{2}\left(\frac{5^{4}}{6}\right) \cdot\left(\frac{1}{6}\right)^{2}$

$=1 \cdot\left(\frac{5}{6}\right)^{6}+6 \cdot \frac{1}{6} \cdot\left(\frac{5}{6}\right)^{5}+15 \cdot \frac{1}{36} \cdot\left(\frac{5}{6}\right)^{4}$

$=\left(\frac{5}{6}\right)^{6}+\left(\frac{5}{6}\right)^{5}+\frac{5}{12} \cdot\left(\frac{5}{6}\right)^{4}$

$=\left(\frac{5}{6}\right)^{4}\left[\left(\frac{5}{6}\right)^{2}+\left(\frac{5}{6}\right)+\left(\frac{5}{12}\right)\right]$

$=\left(\frac{5}{6}\right)^{4} \cdot\left[\frac{25}{36}+\frac{5}{6}+\frac{5}{12}\right]$

$=\left(\frac{5}{6}\right)^{4} \cdot\left[\frac{25+30+15}{36}\right]$

$=\frac{70}{36} \cdot\left(\frac{5}{6}\right)^{4}$

$=\frac{35}{18} \cdot\left(\frac{5}{6}\right)^{4}$

 

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.