An urn contains 3 yellow and 4 green balls. Find the probability distribution of the
number of green balls in three draws when a ball is drawn at random with
replacement. Also find its mean and variance.
Answers
Mean = 12/7 & Variance = 48/49 of Green Balls if 3 Balls are drawn with replacement if An urn contains 3 yellow and 4 green balls
Step-by-step explanation:
An urn contains 3 yellow and 4 green balls
Total Balls = 7
Probability of Green Ball = 4/7 = p
Probability of not green Ball = 1 - p = 1 - 4/7 = 3/7 = q
P(X) = ⁿCₓpˣqⁿ⁻ˣ
n = 3 Draws
P(0) = ³C₀(4/7)⁰(3/7)³ = 27/343
P(1) = ³C₁(4/7)¹(3/7)² = 108/343
P(2) = ³C₂(4/7)²(3/7)¹ = 144/343
P(3) = ³C₃(4/7)³(3/7)⁰ = 64/343
Expected Number of Balls = 0*27/343 + 1 * 108/343 + 2 * 144/343 + 3 *64/343 = 588/343 = 12/7
or
Mean = np = 3 * 4/7 = 12/7
Variance = npq = (12/7)(4/7) = 48/49
Learn more:
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Answer:
variance = 36/49
Step-by-step explanation:
the above explanation is right except variance part
variance = npq
np = mean = 12/7
variance = 12/7 * q
q= 3/7
thus variance = 12/7 * 3/7
variance = 36/49 right answer for the above sum