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A certain coin is tossed with probability of showing head being ‘p’. Let ‘q’ denotes the probability that when the coin is tossed four times, the number of heads obtained is even, then

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There are exactly two values of ‘p’, if $q = \frac{3}{5}$

There is no value of p, if $q = \frac{1}{4}$

There is exactly one value of ‘p’, if $q = \frac{3}{4}$

There are exactly four values of ‘p’, if $q = \frac{4}{5}$

answer is A, C.

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Detailed Solution

$Q = P (0 (or) 2 (or) 4 heads) = p^{4} + 6 p^{2} {(1 - p)}^{2} + {(1 - p)}^{4} = \frac{{(2 p - 1)}^{4} + 1}{2}$

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