In a series of 4 trials, the probability of getting two successes is equal to the probability of getting three successes. The probability of getting at least one success is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{609}{625}\)
Let p be the probability of success in each trial and q = 1 − p. The number of successes X is binomial with n = 4:
\(P(X = r) = {}^{4}C_{r}\,p^r q^{4-r}\)
Given P(X = 2) = P(X = 3):
\(6p^2q^2 = 4p^3q\)
Dividing by \(2p^2q\) (p and q are not 0):
\(3q = 2p\)
\(3(1 - p) = 2p\)
\( p = \frac{3}{5},\quad q = \frac{2}{5}\)
“At least one success” is the complement of “no success”:
\( \displaystyle P(X \ge 1) = 1 - q^4 = 1 - \left(\frac{2}{5}\right)^4 \)
\( \displaystyle = 1 - \frac{16}{625} = \frac{609}{625} \)