Probability 3 Question 44
44. A box contanis 2 black, 4 white and 3 red balls. One ball is drawn at random from the box and kept aside. From the remaining balls in the box, another ball is drawn at random and kept beside the first. This process is repeated till all the balls are drawn from the box. Find the probability that the balls drawn are in the sequence of 2 black, 4 white and 3 red.
$(1979,2 M)$
Integer Answer Type Question
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Answer:
Correct Answer: 44. (b)
Solution:
- Since, the drawn balls are in the sequence black, black, white, white, white, white, red, red and red.
Let the corresponding probabilities be
Then, $\quad p _1=\frac{2}{9}, p _2=\frac{1}{8}, p _3=\frac{4}{7}, p _4=\frac{3}{6}, p _5=\frac{2}{5}$
$$ p _6=\frac{1}{4}, p _7=\frac{3}{3}, p _8=\frac{2}{2}, p _9=1 $$
$\therefore$ Required probabilitie
$$ \begin{aligned} & p _1 \cdot p _2 \cdot p _3 \cdot \ldots \cdot p _9 \\ & =\frac{2}{9} \quad \frac{1}{8} \quad \frac{4}{7} \quad \frac{3}{6} \quad \frac{2}{5} \quad \frac{1}{4} \quad \frac{3}{3} \quad \frac{2}{2} \quad(1)=\frac{1}{1260} \end{aligned} $$
