Question

An unbiased coin is tossed 6 times. The probability that 3rd

head appears on the 6th trial is

• A. $\frac{5}{16}$
• B. $\frac{5}{32}$
• C. $\frac{5}{8}$
• D. $\frac{5}{64}$

Answer 1

Answer: (B)

$\frac{5}{32}$

Answer 2

Prob. that in the first 5 trials 2 are heads

=⁵C₂ $\left( \frac{1}{2} \right)^{2}\left( \frac{1}{2} \right)^{3} = \frac{10}{32} = \frac{5}{16}$

\ Prob. that 3rd head is at 6th = $\frac{5}{16} \times \frac{1}{2} = \frac{5}{32}$