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A bag contains 9 identical black balls numbered 1 to 9 and 4 identical white balls numbered 1 to 4 . If 3 balls are drawn at a time randomly from that bag then the probability of getting at least one white ball is
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2043 Upvotes
Verified Answer
The correct answer is:
$\frac{101}{143}$
Total number of white balls $=4$ balls
Total number of black balls $=9$ balls
Possible number of ways to select 3 balls randomly
$$
\begin{aligned}
& ={ }^4 \mathrm{C}_1 \times{ }^9 \mathrm{C}_2+{ }^4 \mathrm{C}_2 \times{ }^9 \mathrm{C}_1+{ }^4 \mathrm{C}_3 \times{ }^9 \mathrm{C}_0 \\
& =18 \times 8+6 \times 9+4=202 \\
& \mathrm{P}(\text { at least one white ball })=\frac{202}{{ }^{13} \mathrm{C}_3} \\
& \frac{202 \times 6}{13 \times 12 \times 11}=\frac{101}{143}
\end{aligned}
$$
Therefore, option (a) is correct.
Total number of black balls $=9$ balls
Possible number of ways to select 3 balls randomly
$$
\begin{aligned}
& ={ }^4 \mathrm{C}_1 \times{ }^9 \mathrm{C}_2+{ }^4 \mathrm{C}_2 \times{ }^9 \mathrm{C}_1+{ }^4 \mathrm{C}_3 \times{ }^9 \mathrm{C}_0 \\
& =18 \times 8+6 \times 9+4=202 \\
& \mathrm{P}(\text { at least one white ball })=\frac{202}{{ }^{13} \mathrm{C}_3} \\
& \frac{202 \times 6}{13 \times 12 \times 11}=\frac{101}{143}
\end{aligned}
$$
Therefore, option (a) is correct.
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