Answer :
Let B1 be the event of grabbing a black checker at the first draw and let R2 be the event of grabbing a red checker at the second draw.
We are asked for the following probability
[tex]P(B_1\text{ \backslash{}cap B\_2)=}[/tex]Final answer:
The probability of Portia grabbing a black checker and then a red checker depends on whether the events are dependent or independent. In this case, the events are dependent.
Explanation:
To find the probability of Portia grabbing a black checker and instantly grabbing a red checker, we need to consider whether the events are dependent or independent. In this case, the events are dependent because the first checker is not replaced before the second grab.
The probability of grabbing a black checker on the first grab is 9/15, as there are 9 black checkers out of 15 total checkers. After removing the black checker, there are now 14 checkers remaining, with 6 of them being red checkers. Therefore, the probability of grabbing a red checker on the second grab, given that a black checker was grabbed on the first grab, is 6/14.
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