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Mathematics · Probability

JEE Main 2024 — 1 February, Shift 1 — Question 1

A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is:

  1. Option A:

    25\frac{2}{5}

  2. Option B:

    27\frac{2}{7}

    Correct
  3. Option C:

    17\frac{1}{7}

  4. Option D:

    15\frac{1}{5}

Answer: B

Step-by-step solution

P(4 W4 B/2 W2 B)=\mathrm{P}(4 \mathrm{~W} 4 \mathrm{~B} / 2 \mathrm{~W} 2 \mathrm{~B})=

P(4W4B)×P(2W2B/4W4B)P(2W6B)×P(2W2B/2W6B)+P(3W5B)×P(2W2B/3W5B)\frac{P(4 W 4 B) \times P(2 W 2 B / 4 W 4 B)}{P(2 W 6 B) \times P(2 W 2 B / 2 W 6 B)+P(3 W 5 B) \times P(2 W 2 B / 3 W 5 B)}

\qquad +P(6W2B)×P(2W2B/6W2B)+P(6 W 2 B) \times P(2 W 2 B / 6 W 2 B)

=15×4C2×4C28C415×4C2×  4C28C4+15×3C2×  5C28C4+...+15×6C2×  2C28C4=\frac{\frac{1}{5}\times \frac{^{4}{{C}_{2}}{{\times }^{4}}{{C}_{2}}}{^{8}{{C}_{4}}}}{\frac{1}{5}\times \frac{^{4}{{C}_{2}}\times \,{{\,}^{4}}{{C}_{2}}}{^{8}{{C}_{4}}}+\frac{1}{5}\times \frac{^{3}{{C}_{2}}\times \,{{\,}^{5}}{{C}_{2}}}{^{8}{{C}_{4}}}+...+\frac{1}{5}\times \frac{^{6}{{C}_{2}}\times \,{{\,}^{2}}{{C}_{2}}}{^{8}{{C}_{4}}}}

=27=\frac{2}{7}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
1 February, Shift 1
Subject
Mathematics
Chapter
Probability
Topic
Total Probability and Baye's Theorem