A large store places its 15 television sets in a clearance sale unknown to anyone, 5 of the television sets are defective. If a customer tests 3 different television sets selected at random, what is the probability distribution of number of defective television sets in the sample?
Answers
Probability distribution of number of defective television sets in the sample
X P(X)
0 8/27
1 12/27
2 6/27
3 1/27
Given:
- 15 television sets
- 5 of the television sets are defective
- customer tests 3 different television sets selected at random
To Find:
- probability distribution of number of defective television sets in the sample
Solution:
Probability of an event = n(E)/n(S)
n(E) = number of possible outcome of event
n(S) = number of possible sample space outcome
Total TV sets = 15
Defective = 5
probability of Defective p = 5/15 = 1/3
Probability of non defective q = 1 - p = 1 - 1/3 = 2/3
n = 3 ( TV selected for testing)
P(x) = ⁿCₓpˣqⁿ⁻ˣ
x = 0 , 1 , 2 , 3 ( x represent number of defectives )
P(0) = ³C₀(1/3)⁰(2/3)³ = 8/27
P(1) = ³C₁(1/3)¹(2/3)² = 12/27
P(2) = ³C₂(1/3)²(2/3)¹ = 6/27
P(3) = ³C₃(1/3)³(2/3)⁰ = 1/27
Probability distribution of number of defective television sets in the sample
X P(X)
0 8/27
1 12/27
2 6/27
3 1/27
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