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Two drawings, each of 3 balls, are made from a bag containing 5 white and 8 black balls, the balls not being replaced before the second trail. Find the chance that the first drawing will give three whites and the second three black balls.

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Hint: These types of questions involve knowledge of probability. First see the total no. of balls available and then after the first draw see how many are left in total and what are the favorable cases.

No. of white balls= 5
No. of black balls=8
Total no. of balls we have = 8+5=13
Since, we are given that there is no replacement i.e balls once taken out will not be included in the next drawing, so
The probability of getting 3 whites in first drawing will be
P1= \[\dfrac{5}{{13}} \times \dfrac{4}{{12}} \times \dfrac{3}{{11}}\]=\[\dfrac{5}{{143}}\]
Here, \[\dfrac{5}{{13}}\] means we took one ball out of 13 and it’s a white ball out of 5 white balls available, then we are left with 4 white balls and 12 total balls and so on.
Similarly, in the second drawing we are left with 10 balls total, out of which 2 are white balls and 8 are black balls.
Therefore, probability of getting 3 black balls in the second drawing is
P2=\[\dfrac{8}{{10}} \times \dfrac{7}{9} \times \dfrac{6}{8}\]=\[\dfrac{7}{{15}}\]
Since, both are independent events therefore, total probability will be the product of both
i.e P=P1+P2
                     =\[\dfrac{5}{{143}} \times \dfrac{7}{{15}}\]
                     =\[\dfrac{7}{{429}}\].

Note: An independent event means the outcome isn’t affected by another event. A dependent event is affected by the outcome of the second event. Example rolling a dice and flipping a coin.