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Help me with part b.
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tree diag?
Maybe you can't have 1 black and 1 red too?!
yes @Callisto
Hmm You are going to pick two balls only..
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Yes
P(X+Y<2) = P(1 black 1 green) + P(1 black 1 red) + P(2 green) I guess :|
There can't be 1 black and 1 red as 1+1=2 It would be P(1 black 1 green)+P(1red 1 green)+P(all green)
I just need a confirmation that it would be like this :/
Ok..
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so am i right?
I guess yes.. *PS: I'm terrible in probability!
okay what would be the probability like. I mean is P(bg)=(3/8)*(3/8)
No?!
then?
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P(1 Black 1 green) = P(black then green) + P(green then black) P (black then green) = 3/8 * 3/7
how do you know its with or without replacement?
I supposed it is without replacement.
okay thanks.
@Callisto What about part (e)?
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@ash2326
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