Probability Ques 103
- A bag contains 12 red balls and 6 white balls. Six balls are drawn one by one without replacement of which at least 4 balls are white. Find the probability that in the next two drawn exactly one white ball is drawn. (Leave the answer in ${ }^{n} C _r$ ).
$(2004,4$ M)
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Answer:
Correct Answer: 103.$\frac{{ }^{12} C _2 \cdot{ }^{6} C _4}{{ }^{18} C _6} \cdot \frac{{ }^{10} C _1 \cdot{ }^{2} C _1}{{ }^{12} C _2}+\frac{{ }^{12} C _1 \cdot{ }^{6} C _5}{{ }^{18} C _6} \cdot \frac{{ }^{11} C _1 \cdot{ }^{1} C _1}{{ }^{12} C _2} \quad$ 19. $\frac{9 m}{8 N+m}$
Solution:
Formula:
- Let $A _1$ be the event exactly 4 white balls have been drawn. $A _2$ be the event exactly 5 white balls have been drawn.
$A _3$ be the event exactly 6 white balls have been drawn.
$B$ be the event exactly 1 white ball is drawn from two draws. Then,
$$ P(B)=P \frac{B}{A _1} P\left(A _1\right)+P \frac{B}{A _2} P\left(A _2\right)+P \frac{B}{A _3} P\left(A _3\right) $$
But $P \frac{B}{A _3}=0$
[since, there are only 6 white balls in the bag]
$$ \begin{aligned} \therefore \quad P(B) & =P \frac{B}{A _1} P\left(A _1\right)+P \frac{B}{A _2} P\left(A _2\right) \\ & =\frac{{ }^{12} C _2 \cdot{ }^{6} C _4}{{ }^{18} C _6} \cdot \frac{{ }^{10} C _1 \cdot{ }^{2} C _1}{{ }^{12} C _2}+\frac{{ }^{12} C _1 \cdot{ }^{6} C _5}{{ }^{18} C _6} \cdot \frac{{ }^{11} C _1 \cdot{ }^{1} C _1}{{ }^{12} C _2} \end{aligned} $$
BETA
