3 white balls and 5 black balls are placed in a bag and three men draw a ball in succession (the balls drawn are not being replaced), until a white ball is drawn, the ratio of their respective chances is
A. 27 : 18 : 11
B. 11 : 18 : 27
C. 18 : 11 : 27
D. 18 : 27 : 11
Answer
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Hint: Here, we use the concept of probability. Using formula, find the probabilities of drawing white ball by P, Q and R. Keeping in mind their sequence. At last find their ration by comparing all three values. As given that the balls drawn are not being replaced, in this case, favorable outcomes and total outcomes both changes after every draw.
Complete step-by-step answer:
Given, there are 3 white balls and 5 black balls.
Let the people be P, Q and R.
Here, assume cases for P, Q and R
Chance of P: (P picks white) Or (P, Q, R pick black and then P picks white)
\[ = \;\dfrac{3}{8} + \dfrac{5}{8} \times \dfrac{4}{7} \times \dfrac{3}{6} \times \dfrac{3}{5} = \dfrac{{27}}{{56}}\]
Chance of Q: (P picks black and Q picks white) Or (P, Q, R pick black and then P picks black and then Q picks white)
\[ = \;\dfrac{5}{8} \times \dfrac{3}{7} + \dfrac{5}{8} \times \dfrac{4}{7} \times \dfrac{3}{6} \times \dfrac{2}{5} \times \dfrac{3}{4} = \dfrac{{18}}{{56}}\]
Chance of R: (P and Q pick black and R picks white) Or (P, Q, R pick black and then P and Q pick black and then R picks white)
\[ = \;\dfrac{5}{8} \times \dfrac{4}{7} \times \dfrac{3}{6} + \dfrac{5}{8} \times \dfrac{4}{7} \times \dfrac{3}{6} \times \dfrac{2}{5} \times \dfrac{1}{4} \times \dfrac{3}{3} = \dfrac{{11}}{{56}}\]
Therefore, the ratio is \[\dfrac{{27}}{{56}}:\dfrac{{18}}{{56}}:\dfrac{{11}}{{56}}\] = 27 : 18 : 11
So, the correct answer is “Option A”.
Note: In these types of questions, always consider the different cases. Here, P, Q and R will draw the ball in sequence i.e., P not get white ball in his/her first draw, then P, Q and R draw balls and then P again draw the ball. This condition is for Q and R also. Also balls are not replaced, in this case, favourable outcomes and total outcomes also changes. Do the calculation carefully.
Complete step-by-step answer:
Given, there are 3 white balls and 5 black balls.
Let the people be P, Q and R.
Here, assume cases for P, Q and R
Chance of P: (P picks white) Or (P, Q, R pick black and then P picks white)
\[ = \;\dfrac{3}{8} + \dfrac{5}{8} \times \dfrac{4}{7} \times \dfrac{3}{6} \times \dfrac{3}{5} = \dfrac{{27}}{{56}}\]
Chance of Q: (P picks black and Q picks white) Or (P, Q, R pick black and then P picks black and then Q picks white)
\[ = \;\dfrac{5}{8} \times \dfrac{3}{7} + \dfrac{5}{8} \times \dfrac{4}{7} \times \dfrac{3}{6} \times \dfrac{2}{5} \times \dfrac{3}{4} = \dfrac{{18}}{{56}}\]
Chance of R: (P and Q pick black and R picks white) Or (P, Q, R pick black and then P and Q pick black and then R picks white)
\[ = \;\dfrac{5}{8} \times \dfrac{4}{7} \times \dfrac{3}{6} + \dfrac{5}{8} \times \dfrac{4}{7} \times \dfrac{3}{6} \times \dfrac{2}{5} \times \dfrac{1}{4} \times \dfrac{3}{3} = \dfrac{{11}}{{56}}\]
Therefore, the ratio is \[\dfrac{{27}}{{56}}:\dfrac{{18}}{{56}}:\dfrac{{11}}{{56}}\] = 27 : 18 : 11
So, the correct answer is “Option A”.
Note: In these types of questions, always consider the different cases. Here, P, Q and R will draw the ball in sequence i.e., P not get white ball in his/her first draw, then P, Q and R draw balls and then P again draw the ball. This condition is for Q and R also. Also balls are not replaced, in this case, favourable outcomes and total outcomes also changes. Do the calculation carefully.
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