The probability of Heather drawing a red marble first, then a blue marble, and finally a green marble is 32 1 . This is determined by multiplying the probabilities of each individual marble drawn. The option chosen is 32 1 .
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Calculate the probability of drawing a red marble: 4 1 .
Calculate the probability of drawing a blue marble: 4 1 .
Calculate the probability of drawing a green marble: 4 2 = 2 1 .
Multiply the probabilities: 4 1 ⋅ 4 1 ⋅ 2 1 = 32 1 .
Explanation
Understand the problem We are given a bag containing 4 marbles: 1 red, 1 blue, and 2 green. Heather draws one marble, replaces it, and then draws another marble. We want to find the probability of drawing a red marble first, then a blue marble second, and then a green marble.
Calculate probability of red marble The probability of drawing a red marble on the first draw is 4 1 , since there is 1 red marble out of a total of 4 marbles.
Calculate probability of blue marble Since the marble is replaced, the probability of drawing a blue marble on the second draw is 4 1 , since there is 1 blue marble out of a total of 4 marbles.
Calculate probability of green marble Since the marble is replaced again, the probability of drawing a green marble on the third draw is 4 2 = 2 1 , since there are 2 green marbles out of a total of 4 marbles.
Calculate the combined probability The probability of drawing a red marble first, then a blue marble, and then a green marble is the product of the individual probabilities: P ( R then B then G ) = P ( R ) ⋅ P ( B ) ⋅ P ( G ) = 4 1 ⋅ 4 1 ⋅ 4 2 = 64 2 = 32 1 .
State the final answer Therefore, the probability of drawing a red marble, then a blue marble, and then a green marble is 32 1 .
Examples
This type of probability calculation is useful in scenarios like card games, where you want to know the likelihood of drawing specific cards in a particular order. For example, if you're playing a game where you need to draw a certain sequence of colored cards to win, understanding these probabilities can help you make strategic decisions.