8 coins are tossed simultaneously. The probability of getting at least 6 heads isA. $\frac{7}{64}$B. $\frac{57}{64}$C. $\frac{37}{256}$D. $\frac{249}{256}$

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8 coins are tossed simultaneously. The probability of getting at least 6 heads is
A. $\frac{7}{64}$
B. $\frac{57}{64}$
C. $\frac{37}{256}$
D. $\frac{249}{256}$

8 coins are tossed simultaneously. The probability of getting at least 6 heads is
A. $\frac{7}{64}$
B. $\frac{57}{64}$
C. $\frac{37}{256}$
D. $\frac{249}{256}$

Using Bernoulli’s Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . \mathrm{n}$ and $\mathrm{q}=(1-\mathrm{p})$
As the coin is tossed 8 times the total number of outcomes will be $2^{8}$ .
And we know that the favourable outcomes of getting at least 6 heads are,successes will be, getting a head, The probability of success is $\frac{1}{2}$ and of failure is also $\frac{1}{2}$ the probability of getting at least 6 heads is $=$
$\begin{array}{l} P(6)+P(7)+P(8) \\ ={ }^{8} C_{6}\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{2}+{ }^{8} C_{7}\left(\frac{1}{2}\right)^{7}\left(\frac{1}{2}\right)^{1}+{ }^{8} C_{8}\left(\frac{1}{2}\right)^{8}\left(\frac{1}{2}\right)^{0} \\ \Rightarrow \frac{28+8+1}{256}=\frac{37}{256} \end{array}$
Hence, the correct option is c.