Multiple choice

A certain coin is tossed with probability of showing head being 'p'. Let Q denotes probability that when the coin is tossed two times, the number of heads obtained is odd. Then

  1. there is no value of p, if $Q = \frac { 1 } { 4 }$
  2. there is exactly one value of p, if $Q = \frac { 3 } { 4 }$
  3. there is exactly two value of p, if $Q = \frac { 3 } { 5 }$
  4. there is no value of p, if $Q = \frac { 4 } { 5 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Probability of head = p, tail = 1-p. Two tosses: HH (p^2), HT (p(1-p)), TH ((1-p)p), TT ((1-p)^2). Odd number of heads means HT or TH. Q = 2p(1-p). If Q = 4/5, 2p - 2p^2 = 0.8, 2p^2 - 2p + 0.8 = 0, p^2 - p + 0.4 = 0. Discriminant = 1 - 4(0.4) = 1 - 1.6 = -0.6. No real value for p.

AI explanation

When a coin is tossed two times, the probability of getting an odd number of heads, which means exactly one head, is given by the binomial probability 2C1 x p x (1 - p). This gives the equation Q = 2p - 2p^2, which can be written as 2p^2 - 2p + Q = 0. For this quadratic equation to have a valid probability value between 0 and 1, the maximum possible value of Q is 1/2, so there is no value of p that can make Q = 4/5.

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