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Each of the persons $\mathrm{A}$ and $\mathrm{B}$ independently tosses three fair coins. The probability that both of them get the same number of heads is :
$\frac{1}{8}$
$\frac{5}{8}$
$\frac{5}{16}$
$1$
Solution
$C – I \quad\quad'0'$ Head
$\quad\quad\quad\quad\mathrm{T} \mathrm{T} \mathrm{T} \quad\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{3}=\frac{1}{64}$
$C – II \quad\quad '1'$ head
$\quad\quad\quad\quad\mathrm{HT} \mathrm{T} \quad\left(\frac{3}{8}\right)\left(\frac{3}{8}\right)=\frac{9}{64}$
$C – III \quad\quad'2'$ Head
$\quad\quad\quad\quad\mathrm{H} \mathrm{H} \mathrm{T} \quad\left(\frac{3}{8}\right)\left(\frac{3}{8}\right)=\frac{9}{64}$
$C-IV \quad\quad'3'$ Heads
$\quad\quad\quad\quad\mathrm{H} \mathrm{H} \mathrm{H} \quad\left(\frac{1}{8}\right)\left(\frac{1}{8}\right)=\frac{1}{64}$
Total probability $=\frac{5}{16}$
