Question

Consider a binomial random variable X. If the probability the first success of this event occurs on the second trial is 0.25, what's the probability p this event will succeed on each trial?

A) 1/4
B) 1/2
C) 1/8
D) 0.125
E) 1

Answer 1

The probability that the first binomial success occurs on the second trial is given as 0.25. Solving the geometric distribution problem by assuming one failure followed by one success, and after manipulating the resulting quadratic equation, we find that the probability of success on each trial is 1/2 (option B).

Step-by-step explanation:

The geometric distribution is relevant while considering the probability that the first success occurs on the second trial. In this situation, we require one failure followed by one success. Since the first success is on the second trial, we can denote the probability of failure as q, and thus the sequence is: failure (q), success (p). The probability of this occurring is q × p, and we are given that this equals 0.25.

For a binomial random variable, we have p + q = 1, meaning q = 1 - p. Plugging in the values, we get (1 - p) × p = 0.25. Through algebraic manipulation, we can solve for p to find the probability of success on each trial:

(1 - p) × p = 0.25

p2 - p + 0.25 = 0

p2 - p = -0.25

The roots of this quadratic equation are p = 0.5 and p = -0.5. Since a probability cannot be negative, we disregard -0.5 and accept 0.5 as the probability of success on each trial. Hence, the probability p this event will succeed on each trial is 1/2, which corresponds to option B.