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If the probability of two successes is 9 times the probability of 3 successes in 3 trials of a binomial distribution, then the probability of success in each trial is:

Solution

✅ Correct Option: 1

For a binomial distribution with 3 trials, the probability of exactly kk successes is:

P(X=k)=C(n,k)×pk×(1−p)n−kP(X = k) = C(n,k) \times p^k \times (1-p)^{n-k}

where n=3n = 3 is the number of trials and pp is the probability of success in each trial.


For 2 successes in 3 trials:

P(X=2)=C(3,2)×p2×(1−p)1P(X=2) = C(3,2) \times p^2 \times (1-p)^1

C(3,2)=3!2!×1!=3C(3,2) = \dfrac{3!}{2! \times 1!} = 3

P(X=2)=3p2(1−p)P(X=2) = 3p^2(1-p)


For 3 successes in 3 trials:

P(X=3)=C(3,3)×p3×(1−p)0P(X=3) = C(3,3) \times p^3 \times (1-p)^0

C(3,3)=1C(3,3) = 1

(1−p)0=1(1-p)^0 = 1

P(X=3)=p3P(X=3) = p^3


Given that the probability of 2 successes is 9 times the probability of 3 successes:

P(X=2)=9×P(X=3)P(X=2) = 9 \times P(X=3)

3p2(1−p)=9p33p^2(1-p) = 9p^3

3p2−3p3=9p33p^2 - 3p^3 = 9p^3

3p2=12p33p^2 = 12p^3

Dividing both sides by 3p23p^2 (where p≠0p \neq 0):

1=4p1 = 4p

p=14p = \dfrac{1}{4}

Therefore, the probability of success in each trial is 14\dfrac{1}{4}.

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