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Red? ○ It's easier to calculate the probability of getting NO red marbles, and subtract that from 1 (we use the.
Typology: Exercises
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● A jar contains 30 red marbles, 12 yellow marbles, 8 green marbles and 5 blue marbles ● What is the probability that you draw and replace marbles 3 times and you get NO red marbles? ● There are 55 marbles, 25 of which are not red ● (^) P(getting a color other than red) = P(25/55) ≈. ● Probability of this happening 3 times in a row is found by .455.455.455 ≈.
● A jar contains 30 red marbles, 12 yellow marbles, 8 green marbles and 5 blue marbles ● What is the probability that you draw and replace marbles 3 times and you get at least 1 Red? ● It's easier to calculate the probability of getting NO red marbles, and subtract that from 1 (we use the complement rule : P(A C ) = 1 – P(C) ● From previous example, it is 1 - .094 =.
● A jar contains 30 red marbles, 12 yellow marbles, 8 green marbles and 5 blue marbles ● You draw and replace marbles 3 times. What is the probability you draw 1 Red, 1 Yellow, and 1 Blue? ● This is harder, because we are drawing marbles in an order, but we don't care about which order we get Red, Yellow and Blue, just that there is 1 of each. ● But we can do it! ●
● (^) Let RBY = “Draw a Red, then Blue, then Yellow” ● So all disjoint events we want to consider are: RBY, RYB, YRB, YBR, BYR, BRY – there are 6 of them. ● (^) P(RBY) = P(R)P(B)P(Y) = (30/55)(5/55)(12/55) =. ● But we have 6 disjoint cases. Because each one is calculated as a product of the three, and each disjoint case has the same probability (each order is equally likely), our answer is 6*.0108 =.
