A coin is tossed 5 times. Find the probability of getting (i) at least 4 heads,and (ii) at most 4 heads.

 

 

 

 
 
 
 
 

Answers (1)

given: Total number of probability of tossing a coin 5 terms is 32
(i) Probability (at least 4 heads)\Rightarrow p\left ( x= 4 \right )+p\left ( x= 5 \right )
\Rightarrow ^{5}C_{4}\left ( \frac{1}{2} \right )^{1}\left ( \frac{1}{2} \right )^{4}+^{5}C_{5}\left ( \frac{1}{2} \right )^{0}\left ( \frac{1}{2} \right )^{5}
\Rightarrow ^{5}C_{4}\left ( \frac{1}{2} \right )^{5}+\: ^{5}C_{5}\left ( \frac{1}{2} \right )^{5}
\Rightarrow \frac{6}{32}= \frac{3}{16}
(ii) p(atmost 4 heads)= p(x=0)+p\left ( x= 1 \right )+p\left ( x= 2 \right )+p\left ( x= 3 \right )+p\left ( x= 4 \right )
^{5}C_{0}\left ( \frac{1}{2} \right )^{5}+^{5}C_{1}\left ( \frac{1}{2} \right )^{5}+^{5}C_{2}\left ( \frac{1}{2} \right )^{5}+ ^{5}C_{3}\left ( \frac{1}{2} \right )^{5}+^{5}C_{4}\left ( \frac{1}{2} \right )^{5}
\Rightarrow \left ( \frac{1}{2} \right )^{5}\left [1+ 5+10+10+5 \right ]
\Rightarrow \frac{31}{32}

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