Question

A bag contains 5 white marbles, 3 black marbles and 2 green marbles. In each draw, a marble is drawn from the bag and not replaced. In three draws, find the probability of obtaining white,black and green in that order

Answer 1

Hi there!
Here's the answer:

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Given,
No. of White Marbles = 5
No. of Black Marbles = 3
No. of Green Marbles = 2
Total Balls = 10

Given that balls are drawn without replacement
=> No. of balls Left(In each type as well as in total) after drawing each ball is to be considered.

Let E be the Event that in 3 draws white,black & green marbles are obtained in the same order and without replacement.

^^^ Since favourable cases are different, there is no need to track each type of marbles left after a particular draw. But Total No of marbles left is to be noted.

To find P(E):

¶ Probability of Drawing White ball P(W):

• $P(W) = \frac{5C1}{10C1}$

=> $P(W) = \frac{5}{10}$

=> $P(W) = \frac{1}{2}$

Now, There are 9 Left in total

¶ Probability of Drawing Black ball P(B):

• $P(B) = \frac{3C1}{9C1}$

=> $P(B) = \frac{3}{9}$

=> $P(B) = \frac{1}{3}$

Now, there are 8 left in total

¶ Probability of Drawing Green ball P(G):

• $P(G) = \frac{2C1}{8C1}$

=> $P(G) = \frac{2}{8}$

=> $P(G) = \frac{1}{4}$

Now,
Required Probability P(E) = P(W) × P(B) × P(G)

=> $P(E) = \frac{1}{2} × \frac{1}{3} × \frac{1}{4}$

=> $P(E) = \frac{1}{24}$

•°• $\underline{\underline{Required\: Probability = \frac{1}{24}}}$

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:)

Hope it helps