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Question

Mathematics Question on Probability

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

27565\frac{275}{6^5}

B

3654\frac{36}{5^4}

C

18155\frac{181}{5^5}

D

4664\frac{46}{6^4}

Answer

4664\frac{46}{6^4}

Explanation

Solution

The correct answer is (D) : 4664\frac{46}{6^4}
Let probability of getting head = p
So, 5C4p4(1−p)=5C5p5^5C_4p^4(1−p)=^5C_5p^5
⇒p=5(1−p)⇒p=56⇒p=5(1−p)⇒p=\frac{5}{6}
Probability of getting atmost two heads =
5C0(1−p)5+5C1p(1−p)4+5C2p2(1−p)3^5C_0(1−p)^5+^5C_1p(1−p)^4+^5C_2p^2(1−p)^3
=(1+25+250)65=\frac{(1+25+250)}{6^5}
=27665=\frac{276}{6^5}

=4664=\frac{46}{6^4}