Multiple choice

An urn contains $3$ white and $4$ black balls. Three balls are drawn from the urn. The probability distribution function of the number of white ball is x $0$ $1$ $2$ $3$ p(x) $\dfrac{a}{35}$ $\dfrac{b}{35}$ $\dfrac{c}{35}$ $\dfrac{d}{35}$ then values of a.,b,c,d are

  1. $a=1,b=3,c=4,d=5$
  2. $a=4,b=18,c=12,d=1$
  3. $a=4,b=12,c=18,d=2$
  4. $a=2,b=12,c=18,d=2$
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B Correct answer
Explanation

The number of white balls follows a hypergeometric distribution. The total ways to choose 3 balls from 7 is 7C3 = 35. Calculating probabilities for 0, 1, 2, and 3 white balls gives 4/35, 18/35, 12/35, and 1/35 respectively.

AI explanation

The urn contains 7 balls total, so the number of ways to draw 3 balls is 7C3, which equals 35. For 0, 1, 2, and 3 white balls, the respective combinations are 4C3, 3C1 times 4C2, 3C2 times 4C1, and 3C3. Evaluating these gives 4, 18, 12, and 1, making the respective probabilities 4/35, 18/35, 12/35, and 1/35. Thus, a equals 4, b equals 18, c equals 12, and d equals 1.

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