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Adam has a bag containing four yellow gumdrops and three red gumdrops. He will eat one of the gumdrops, and a few minutes later, he will eat a second gumdrop.
1) What is the probability that Adam will eat a yellow gumdrop first and a red gumdrop second?
2) What is the probability that Adam will eat two yellow gumdrops one after the other?
Let me see.
Part 1
A = probability of eating one gumdrop.
B = probability of eating one red gumdrop.
P(A) = 4/7
P(B) = 3/6
P(A n B) = P(A) • P(B)
P(A n B) = (4/7) • (3/6)
Is this right?
Part 2
A = probability of eating one yellow gumdrop.
B = probability of eating another yellow gumdrop.
P(A) = 4/7
P(B) = 3/6
P(A n B) = P(A) • P(B|A)
P(A n B) = (4/7) • (3/6)
3) What is the probability that Adam will eat two gumdrops with the same color?
I don't know how to set this one up. Any ideas?
4) What is the probability that Adam will eat two gumdrops of different colors?
Can someone set this one up for me?
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Parts 1 and 2 correct.
You've done 2 yellows. Find also 2 reds, then add the two answers to get P(two same colour)
Last part: Add P(red then yellow) to P(yellow then red)
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Parts 1 and 2 correct.
You've done 2 yellows. Find also 2 reds, then add the two answers to get P(two same colour)
Last part: Add P(red then yellow) to P(yellow then red)
Bob
Cool. I can do that later. Thank you.
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