MCQEasyJEE 2026Probability Basics

JEE Mathematics 2026 Question with Solution

Bag A contains 99 white and 88 black balls, while bag B contains 66 white and 44 black balls. One ball is picked from B and put in A. Then a ball is drawn from A. Probability it is white is p/qp/q. Find p+qp+q.

  • A

    2323

  • B

    2222

  • C

    2121

  • D

    2424

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Bag A contains 99 white and 88 black balls, and bag B contains 66 white and 44 black balls. One ball is transferred from B to A, then one ball is drawn from A.

Find: The value of p+qp+q if the probability of drawing a white ball from A is p/qp/q.

Use the total probability theorem.

Probability of transferring white from B:

P(TW)=610=35P(T_W)=\frac{6}{10}=\frac{3}{5}

Probability of transferring black from B:

P(TB)=410=25P(T_B)=\frac{4}{10}=\frac{2}{5}

If a white ball is transferred, bag A becomes 10W,8B10W, 8B, so

P(WTW)=1018=59P(W\mid T_W)=\frac{10}{18}=\frac{5}{9}

If a black ball is transferred, bag A becomes 9W,9B9W, 9B, so

P(WTB)=918=12P(W\mid T_B)=\frac{9}{18}=\frac{1}{2}

Therefore,

P(W)=P(WTW)P(TW)+P(WTB)P(TB)P(W)=P(W\mid T_W)P(T_W)+P(W\mid T_B)P(T_B)

Substituting,

P(W)=59×35+12×25=13+15=815P(W)=\frac{5}{9}\times \frac{3}{5}+\frac{1}{2}\times \frac{2}{5}=\frac{1}{3}+\frac{1}{5}=\frac{8}{15}

So p=8p=8 and q=15q=15. Hence,

p+q=8+15=23p+q=8+15=23

Therefore, the correct option is A.

Common mistakes

  • Students may directly use the original composition of bag A and take the probability as 917\frac{9}{17}. This is wrong because one transfer happens before the draw. First update bag A based on the transferred ball, then compute the final probability.

  • Students may forget to use conditional probability and average the cases incorrectly. The two transfer cases are not equally likely. Use weighted probabilities: multiply each conditional probability by the probability of that transfer case.

  • Students may make the denominator of bag A after transfer as 1717 instead of 1818. This is wrong because one extra ball is added to bag A before drawing. After transfer, bag A always contains 1818 balls.

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