The answers are a) .3403 b) .1224 and c) .4737. Please explain how to solve the
ID: 2947703 • Letter: T
Question
The answers are a) .3403 b) .1224 and c) .4737. Please explain how to solve the problem step by step.
Box A has 3 green marbles and 1 red marble, box B has 4 green marbles and 2 red marbles, and box C has 5 green marbles and 3 red marbles. Peter rolls a die first. If the die comes up 1, then he will randomly choose a marble from box A; if the die comes up 2 or 3, he will randomly choose a marble from box B; if the die comes up 4, 5 or 6, he will randomly choose a marble from box C (a) What is the probability that Peter gets a red marble? (b) If Peter gets a red marble, what is the probability that the marble is chosen from box A? (c) If Peter gets a green marble, what is the probability that the marble is chosen from box C?Explanation / Answer
The number of differently coloured marbles in the 3 boxes and the probability of selecting a box can be given here as:
a) Probability that a red marble is selected is computed here as:
= (1/6)*(1/4) + (2/6)*(2/6) + (3/6)*(3/8)
= (1/24) + (1/9) + (3/16)
= 0.3403
Therefore 0.3403 is the required probability here.
b) Given that a red marble is drawn, probability that it is from box A is computed as:
= (1/6)*(1/4) / P( red )
= (1/24) / 0.3403
= 0.1224
Therefore 0.1224 is the required probability here.
c) Again we first compute here:
P( green ) = (1/6)*(3/4) + (2/6)*(4/6) + (5/8)*(3/6) = 0.6597
Now given that a green is taken out, probability that it is taken out from C is computed as:
= (5/8)*(3/6) / 0.6597
= 0.4737
Therefore 0.4737 is the required probability here.
Box A B C Green Marbles 3 4 5 Red Marbels 1 2 3 Probability of that box 1/6 2/6 3/6Related Questions
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