SOLUTION: You pick two marbles from a bag. You pick the second one without replacing the first marble. (2 green, 2 white, 2 red, 2 yellow, and 2 blue)
P(red then blue) =
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-> SOLUTION: You pick two marbles from a bag. You pick the second one without replacing the first marble. (2 green, 2 white, 2 red, 2 yellow, and 2 blue)
P(red then blue) =
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Question 1163938: You pick two marbles from a bag. You pick the second one without replacing the first marble. (2 green, 2 white, 2 red, 2 yellow, and 2 blue)
P(red then blue) =
P(white then blue then white) =
(I'm really struggling! Please help me if you can) Found 2 solutions by solver91311, ikleyn:Answer by solver91311(24713) (Show Source):
The answer to your second question is easy. Since you only picked two marbles, the probability of getting a white then a blue and then another white (or a third marble of any color for that matter) is zero.
The probability of a red on the first draw is 2/10 or 1/5. But since you did not replace the first marble drawn, when you go to pick the second one there are still two blue ones but only nine total marbles, so your probability of a blue on the second draw given any other color on the first draw is 2/9.
Since you have accounted for the change in the number of marbles from which to draw for the second draw, the two events are independent and the multiplication rule applies.
John
My calculator said it, I believe it, that settles it
