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}

 

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