JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Probability

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Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is:

(A) $\frac{275}{6 ^{5}}$
(B) $\frac{36}{5 ^{4}}$
(C) $\frac{181}{5 ^{5}}$
(D) $\frac{46}{6 ^{4}}$ ✓ Correct

Answer: (D)

Solution

Let

P (H) = x

and

P (T) = 1 - x

. Given

P (4 H, 1 T) = P (5 H)

^{5} C_{1} x^{4} (1 - x) =^{5} C_{5} x^{5}

5 (1 - x) = x

x = \frac{5}{6}

Probability of at most 2 heads:

P (0 H) + P (1 H) + P (2 H)

=^{5} C_{0} (\frac{1}{6})^{5} +^{5} C_{1} \frac{5}{6} (\frac{1}{6})^{4} +^{5} C_{2} (\frac{5}{6})^{2} (\frac{1}{6})^{3}

= \frac{1}{6 ^{5}} (1 + 25 + 250) = \frac{276}{6 ^{5}} = \frac{46}{6 ^{4}}

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Single correct option

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