for a binomial distribution mean is 4 and variance is 2. find probability of getting atleast 2 successes.
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Mean = n p = 4
Variance = n p q = 2
Dividing them, q = 1/2
So, p = 1/2
So, n = 8
Probability of getting atleast two successes = 1 - {P(x = 0) + P(x = 1) }
P(x = 0) = 8 C0 / 2^(8) = 1/2^(8)
P(x = 1) = 8C1/ (2^8) = 8 /2^(8)
P(x = 0) + P(x = 1) = 9/2^(8) = 9/256
Probability of getting atleast two successes = 1 - (9/256) = 247 / 256
Variance = n p q = 2
Dividing them, q = 1/2
So, p = 1/2
So, n = 8
Probability of getting atleast two successes = 1 - {P(x = 0) + P(x = 1) }
P(x = 0) = 8 C0 / 2^(8) = 1/2^(8)
P(x = 1) = 8C1/ (2^8) = 8 /2^(8)
P(x = 0) + P(x = 1) = 9/2^(8) = 9/256
Probability of getting atleast two successes = 1 - (9/256) = 247 / 256
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