An urn contains a total of 12 chocolates, out of which 4 are milk and the remaining are dark. Suppose a sample of 5 chocolates is drawn from the urn one by one (with replacement). What is the conditional probability that the first, third, and fifth chocolates will be dark given that exactly 4 chocolates out of the 5 in the sample are dark? Enter the answer upto 1 decimal accuracy.
Answers
Answer:
32/243
Step-by-step explanation:
possibilities:
11110
11101
11011
10111
01111
favourable: 2nd and 4th
Given : An urn contains a total of 12 chocolates, out of which 4 are milk and the remaining are dark.
a sample of 5 chocolates is drawn from the urn one by one (with replacement)
To Find : conditional probability that the first, third, and fifth chocolates will be dark given that exactly 4 chocolates out of the 5 in the sample are dark
Solution:
Total chocolates = 12
Milk = 4
Dark = 12 - 4 = 8
Probability of milk chocolate = 4/12 = 1/3
Probability of dark chocolate = 8/12 = 2/3
sample of 5 with replacement
4 chocolates out of the 5 in the sample are dark
=> P(4) = ⁵C₄(2/3)⁴(1/3)¹ = 80/243
first, third, and fifth chocolates will be dark
2nd chocolate milk or 4th chocolate milk
= (2/3)(1/3)(2/3)(2/3)(2/3) + (2/3)(2/3)(2/3)(1/3)(2/3)
= 16/243 + 16/243
= 32/243
(32/243 )/ (80/243)
= 32/80
= 2/5
= 0.4
Conditional probability that the first, third, and fifth chocolates will be dark given that exactly 4 chocolates out of the 5 in the sample are dark = 0.4
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