Multiple choice

A biased coin is tossed twice.The probability of head is twice the tail.The PDF of number of heads is x $0$ $1$ $2$ p(x) $\dfrac{a}{d}$ $\dfrac{b}{d}$ $\dfrac{c}{d}$ then values of $a,b,c,d$ are

  1. $a=1,b=2,c=3,d=4$
  2. $a=1,b=4,c=4,d=9$
  3. $a=1,b=4,c=4,d=10$
  4. $a=1,b=2,c=1,d=4$
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B Correct answer
Explanation

Let P(Tail) = p, then P(Head) = 2p. p + 2p = 1 => p = 1/3. P(H) = 2/3, P(T) = 1/3. For two tosses: P(0 heads) = P(TT) = 1/3 * 1/3 = 1/9. P(1 head) = P(HT, TH) = 2 * (2/3 * 1/3) = 4/9. P(2 heads) = P(HH) = 2/3 * 2/3 = 4/9. Thus a=1, b=4, c=4, d=9.

AI explanation

Using the given probability distribution, the probability of a tail is 1/3 and the probability of a head is 2/3. The probability of getting 0 heads is (1/3) squared, which is 1/9. The probability of getting exactly 1 head is 2 multiplied by (1/3) multiplied by (2/3), which equals 4/9, and the probability of getting 2 heads is (2/3) squared, which equals 4/9. This gives values of a=1, b=4, c=4, and d=9.

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