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A coin is tossed 5 times. What is the probability of gettiing (i) 3 heads, (ii) at most 3 heads?

Answers (1)

Here $n=5,p=\frac{1}{2}\: \: and\: \: q=\frac{1}{2}$

we know that,

$p(x)=_{}^{5}\textrm{C}_3$ $\left ( \frac{1}{2} \right )^3\left ( \frac{1}{3} \right )^{5-3}$

$=_{}^{5}\textrm{C}_3$ $\left ( \frac{1}{2} \right )^3\left ( \frac{1}{3} \right )^2$

$=_{}^{5}\textrm{C}_3$ $\left ( \frac{1}{2} \right )^5$

$\rightarrow \frac{5\times 4\times3}{2\times3}\left ( \frac{1}{2} \right )^5\Rightarrow10\left ( \frac{1}{2} \right )^5$

$\Rightarrow \frac{5}{16}$

(ii) Probability of getting at most 3 heads is

$P\left ( x\leq 3 \right )=P(0)+P(1)+P(2)+P(3)$

Then

$P(0)=_{}^{5}\textrm{C}_0$ $\left ( \frac{1}{2} \right )^5$

$=\frac{1}{32}$

$P(1)=_{}^{5}\textrm{C}_1$ $\left ( \frac{1}{2} \right )^5=\frac{5}{32}$

$P(2)=_{}^{5}\textrm{C}_2$ $\left ( \frac{1}{2} \right )^5=\frac{5}{16}$

$P(3)=_{}^{5}\textrm{C}_3$ $\left ( \frac{1}{2} \right )^5=\frac{5}{16}$

$\therefore P(x\leq 3)=\frac{1}{32}+\frac{5}{32} + \frac{5}{16} + \frac{5}{16}=\frac{26}{32}$

$=\frac{13}{16}$

Posted by

Ravindra Pindel

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