The probability that the fruits will be selected in the order of banana then apple then orange is 0.1094
Step-by-step explanation:
The given information are;
The number of bananas = 4
The number of apples = 3
The number of oranges = 5
The total number of fruits are 4 + 3 + 5 = 12
The probability of selecting a banana, p(b) = 4/12 = 1/3
The probability of selecting an apple, p(a) = 3/12 = 1/4
The probability of selecting an oranges, p(o) = 5/12
The probability of selecting one banana, then one apple, then one orange, in that order, from the basket with replacement is the product of the individual probabilities found as follows;
p(p(b) then p(a) then p(o)) =
The number o
The probability that the first fruit is a banana = ₁₁C₃×(1/3)^(3)×(2/3)^8 = 0.2384
The probability that the second fruit is an apple = ₁₁C₃×(1/4)^(3)×(3/4)^8 = 0.258
The probability that the third fruit is an apple = ₁₁C₃×(5/12)^(3)×(7/12)^8 = 0.16
The total probability = 0.2384 + 0.285 + 0.16 = 0.657
The number of ways in which the three can be selected = 3 × 2 × 1 = 6 ways
Therefore, the probability that the fruits will be selected in the order of banana then apple then orange = 0.657/6 = 0.1094.
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