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 :

<p>Coin is tossed 5 times, so n = 5</p> <p>Let, p = probability of getting heads</p> <p>q = probability of getting tails.</p> <p>$\therefore$ p + q = 1 ...... (1)</p> <p>$\therefore$ Probability of getting 4 heads</p> <p>= ⁵C₄ . p⁴ . q</p> <p>And probability of getting 5 heads</p> <p>= ⁵C₅ . p⁵</p> <p>Given, ⁵C₄ . p⁴ . q = ⁵C₅ . p⁵</p> <p>$\Rightarrow$ 5q = p ....... (2)</p> <p>From equation (1) and (2), we get,</p> <p>5q + q = 1</p> <p>$\Rightarrow$ 6q = 1</p> <p>$\Rightarrow$ q = ${1 \over 6}$</p> <p>$\therefore$ p = 1 $-$ ${1 \over 6}$ = ${5 \over 6}$</p> <p>Now, probability of getting atmost two heads</p> <p>= p (x = 0) + p (x = 1) + p (x = 2)</p> <p>p (x = 0) = Getting zero head in 5 trials</p> <p>= ⁵C₀ . p⁰ . q⁵</p> <p>p (x = 1) = Getting one head in 5 trials</p> <p>= ⁵C₁ . p¹ . q⁴</p> <p>p (x = 2) = Getting two heads in 5 trials</p> <p>= ⁵C₂ . p² . q³</p> <p>= ⁵C₀ . q⁵ + ⁵C₁ . pq⁴ + ⁵C₂ . p²q³</p> <p>$$ = {\left( {{1 \over 6}} \right)^5} + 5\,.\,{5 \over 6}\,.\,{\left( {{1 \over 6}} \right)^4} + 10\,.\,{\left( {{5 \over 6}} \right)^2}\,.\,{\left( {{1 \over 6}} \right)^3}$$</p> <p>$= {{1 + 25 + 250} \over {{6^5}}} = {{276} \over {{6^5}}}$ = ${{46} \over {{6^4}}}$