Find the probability that in five tosses of a fair die a 3 appears at most once
Answers
Answer:3125/3888
Step-by-step explanation:
Probability of obtaining a 3 at most once= 5(1/6*5/6*5/6*5/6*5/6) + (5/6)^5 = 5(625/7776) + 3125/7776
= 3125/7776 + 3125/7776
= 3125/3888
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Given : five tosses of a fair die
To Find : Probability that 3 will appear
at most once
Solution:
n = 5
p = 1/6
q = 1 - p = 1 - 1/6 = 5/6
P(x) = ⁿCₓpˣqⁿ⁻ˣ
Probability that 3 will appear
at most once
= P(0) + P(1)
= ⁵C₀(1/6)⁰(5/6)⁵ + ⁵C₁(1/6)¹(5/6)⁴
= (5⁵ + 5⁵)/6⁵
= 6250/7776
= 0.803755
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