A basket of fruit contains 4 bananas, 3 apples, and 5 oranges. You intend to draw a piece of fruit from the basket, keep it, draw a 2nd piece of fruit from the basket, keep that, and then draw a third piece of fruit from the same basket. What is the probability of selecting one banana, then one apple, then one orange, in that order, from the basket if you are blindfolded?

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.

any options?

step-by-step explanation:

on monday rodney ate 1/4th of the sandwich = 3/12

on tuesday he ate 1/3rd of the sandwich = 4/12

therefore in total he has eaten = 3/12 + 4/12 = 7/12

thus we can say that he has eaten 7/12 of the sandwich on monday and tuesday alltogether.

happy to

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