Question

A coin is tossed 10 times. The probability of getting exactly six heads is

(A) \({ }^{10} C_{6}\left(\frac{1}{2}\right)^{6}\)

(B) \({ }^{10} C_{6}\left(\frac{1}{2}\right)^{7}\)

(C) \({ }^{10} C_{6}\left(\frac{1}{2}\right)^{8}\)

(D) \({ }^{10} C_{6}\left(\frac{1}{2}\right)^{10}\)

Answer 1

Correct option is (D) \({ }^{10} C_{6}\left(\frac{1}{2}\right)^{10}\)

If a coin is tossed then the probability of getting head and getting tail is \(\frac 12\).

Now, if the coin is tossed 10 times then the probability of getting exactly 6 heads is,

\({ }^{10} C_{6} \times \left(\frac{1}{2}\right)^{6}\times \left(\frac{1}{2}\right)^{10 -6}\)

\(={ }^{10} C_{6} \times \left(\frac{1}{2}\right)^{6}\times \left(\frac{1}{2}\right)^{4}\)

\(={ }^{10} C_{6}\left(\frac{1}{2}\right)^{10}\)