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A box contains 20 identical balls of which 10 are blue and 10 are green. The balls are drawn at random from the box one at a time with replacement. The probability that a blue ball is drawn 4th time on the 7 th draw is
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The correct answer is:
$\frac{5}{32}$
Probability of getting a blue ball at any draw
$$
=\mathrm{p}=\frac{10}{20}=\frac{1}{2}
$$
$\mathrm{P}$ [getting a blue ball 4th time in 7 th draw] $=\mathrm{P}$ [getting 3 blue balls in $6 \mathrm{draw}] \times \mathrm{P}[\mathrm{a}$ blue ball in the 7 th draw].
$$
\begin{array}{l}
={ }^{6} \mathrm{C}_{3}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{3} \cdot \frac{1}{2} \\
=\frac{6 \times 5 \times 4}{1 \times 2 \times 3}\left(\frac{1}{2}\right)^{7}=20 \times \frac{1}{32 \times 4}=\frac{5}{32}
\end{array}
$$
$$
=\mathrm{p}=\frac{10}{20}=\frac{1}{2}
$$
$\mathrm{P}$ [getting a blue ball 4th time in 7 th draw] $=\mathrm{P}$ [getting 3 blue balls in $6 \mathrm{draw}] \times \mathrm{P}[\mathrm{a}$ blue ball in the 7 th draw].
$$
\begin{array}{l}
={ }^{6} \mathrm{C}_{3}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{3} \cdot \frac{1}{2} \\
=\frac{6 \times 5 \times 4}{1 \times 2 \times 3}\left(\frac{1}{2}\right)^{7}=20 \times \frac{1}{32 \times 4}=\frac{5}{32}
\end{array}
$$
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