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Let X be the number of successes in 'n' independent Bernoulli trials with probability of success p = ¾, The least value of ‘n’ so that P(X ≥ 1) ≥ 0.9375 is .......
- a)2
- b)1
- c)4
- d)3
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Let X be the number of successes in 'n' independent Bernoulli trials ...
Explanation:
Concept:
In a Bernoulli trial, the probability of success is denoted by 'p' and the probability of failure is denoted by 'q' where q = 1 - p. The probability mass function of the number of successes 'X' in 'n' trials is given by the binomial distribution formula P(X = k) = nCk * p^k * q^(n-k).
Finding the least value of 'n':
We need to find the least value of 'n' such that P(X ≥ 1) ≥ 0.9375.
P(X ≥ 1) = 1 - P(X = 0) = 1 - (1 - p)^n ≥ 0.9375
=> (1 - p)^n ≤ 0.0625
=> (1 - 3/4)^n ≤ 0.0625
=> (1/4)^n ≤ 1/16
=> 4^n ≤ 16
=> n ≥ 2
Answer:
Therefore, the least value of 'n' so that P(X ≥ 1) ≥ 0.9375 is 2. Hence, the correct answer is option 'A'.
Concept:
In a Bernoulli trial, the probability of success is denoted by 'p' and the probability of failure is denoted by 'q' where q = 1 - p. The probability mass function of the number of successes 'X' in 'n' trials is given by the binomial distribution formula P(X = k) = nCk * p^k * q^(n-k).
Finding the least value of 'n':
We need to find the least value of 'n' such that P(X ≥ 1) ≥ 0.9375.
P(X ≥ 1) = 1 - P(X = 0) = 1 - (1 - p)^n ≥ 0.9375
=> (1 - p)^n ≤ 0.0625
=> (1 - 3/4)^n ≤ 0.0625
=> (1/4)^n ≤ 1/16
=> 4^n ≤ 16
=> n ≥ 2
Answer:
Therefore, the least value of 'n' so that P(X ≥ 1) ≥ 0.9375 is 2. Hence, the correct answer is option 'A'.
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Community Answer
Let X be the number of successes in 'n' independent Bernoulli trials ...
We have, p = ¾, q = 1 - p = ¼
It is given that P (X ≥ 1) ≥ 0.9375
= 1 - P(X = 0) ≥ 0.9375
= 1 - nCo(po)(g)n-o ≥ 0.9375
= 1 - (¼)n ≥ 0.9375
= 1 - 0.9375 ≥ (¼)n= 0.0625 ≥ (¼)n
= 625/10000 ≥ (¼)n
= 1/16 ≥ (¼)n
= 16 ≤ 4th
⇒ n = 2
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