Re: A jar contains 3 red and 2 white marbles. Two marbles are picked witho
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19 Mar 2023, 11:51
OE
The easiest way to compute the probability in question is through the “1 − x” shortcut. To do so, imagine the opposite of the event of interest, namely, that none of the n throws yields a 6. The probability of a single throw not yielding a 6 is 5/6, and because each throw is independent, the cumulative probability of none of the n throws yielding a 6 is found by multiplication:
P(RR)=3/5*2/4=3/10
Next, consider the probability of picking a red marble followed by a white marble:
P(RW)=3/5*2/4=3/10
However, this is not the only way to pick one red AND one white marble; you could have picked the white one first, followed by the red one:
P(WR)=2/5*3/4=3/10
This event is mutually exclusive from picking a red marble followed by a white marble. Thus, the total probability of picking one red AND one white marble is the sum of the probabilities of RW and WR, yielding an answer of:
P(RW or WR)=3/10+3/10=3/5
A = 3/10
B= 3/5
B>A