A die is tossed \displaystyle{3}3 times. What is the probability of (a) No fives turning up? (b) \displaystyle{1}1 five? (c) \displaystyle{3}3 fives?

A die is tossed $\displaystyle{3}$ times. What is the probability of

(a) No fives turning up?

(b) $\displaystyle{1}$ five?

(c) $\displaystyle{3}$ fives?

Solution

 This is a binomial distribution because there are only 

$\displaystyle{2}$ possible outcomes (we get a $\displaystyle{5}$ or we don't).

Now, $\displaystyle{n}={3}$ for each part. Let $\displaystyle{X}=$ number of fives appearing.

(a) Here, x = 0.

$\displaystyle{P}{\left({X}={0}\right)}$ $\displaystyle={{C}_{{x}}^{{n}}}{p}^{x}{q}^{{{n}-{x}}}$ $\displaystyle={{C}_{{0}}^{{3}}}{\left(\frac{1}{{6}}\right)}^{0}{\left(\frac{5}{{6}}\right)}^{3}$ $\displaystyle=\frac{125}{{216}}$ $\displaystyle={0.5787}$

(b) Here, x = 1.

$\displaystyle{P}{\left({X}={1}\right)}$ $\displaystyle={{C}_{{x}}^{{n}}}{p}^{x}{q}^{{{n}-{x}}}$ $\displaystyle={{C}_{{1}}^{{3}}}{\left(\frac{1}{{6}}\right)}^{1}{\left(\frac{5}{{6}}\right)}^{2}$ $\displaystyle=\frac{75}{{216}}$ $\displaystyle={0.34722}$

(c) Here, x = 3.

$\displaystyle{P}{\left({X}={3}\right)}={{C}_{{x}}^{{n}}}{p}^{x}{q}^{{{n}-{x}}}$ $\displaystyle={{C}_{{3}}^{{3}}}{\left(\frac{1}{{6}}\right)}^{3}{\left(\frac{5}{{6}}\right)}^{0}$ $\displaystyle=\frac{1}{{216}}$ $\displaystyle={4.6296}\times{10}^{ -{{3}}}$