10 coins are tossed simultaneously. Find the probability of getting(i) exactly 3 heads(ii) not more than 4 heads

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10 coins are tossed simultaneously. Find the probability of getting
(i) exactly 3 heads
(ii) not more than 4 heads

10 coins are tossed simultaneously. Find the probability of getting
(i) exactly 3 heads
(ii) not more than 4 heads

(i) As 10 coins are tossed simultaneously the total number of outcomes are $2^{10}=1024$ . the favourable outcomes of getting exactly 3 heads will be
${ }^{10} \mathrm{C}_{3}=120$
Thus, the probability
$\begin{array}{l} =\frac{\text { The favourable outcomes }}{\text { The total number of outcomes }} \\ =\frac{120}{1024} \\ =\frac{15}{128} \end{array}$
Hence, the probability is $\frac{15}{128}$ .
(ii) As 10 coins are tossed simultaneously the total number of outcomes are $2^{10}=1024$ .
the favourable outcomes of getting not more than 4 heads will be
${ }^{10} \mathrm{C}_{0}+{ }^{10} \mathrm{C}_{1}+{ }^{10} \mathrm{C}_{2}+{ }^{10} \mathrm{C}_{3}+{ }^{10} \mathrm{C}_{4}=386$
Thus, the probability
$=\frac{\text { The favourable outcomes }}{\text { The total number of outcomes }}$
$\begin{array}{l} =\frac{386}{1024} \\ \Rightarrow \frac{193}{512} \end{array}$
Hence, the probability is $\frac{193}{512}$ .