A bag contains 5 red and 4 white balls. Four balls are drawn out one by one and not replaced. What is the probability that they are alternatively of different colours?
Answers
Answer:
5/9 is the probability of red balls
4/9 is the probability of white ball
Answer:
First Red = (5/12)
Second white = (7/11)
Third Red = (4/10) { As you have already withdrawn one red ball}
Fourth White = ( 6/9)
Hence overall probability = (5/12)*(7/11)*(4/10)*(6/9) = (7/99)
W R W R : You withdraw white first and then red , white and finally red. The probability:
White first = (7/12)
Second Red = ( 5/11)
Third White = (6/10)
Fourth Red = (4/9)
Hence overall probability= (7/12)*(5/11)*(6/10)*(4/9) = (7/99)
The total probability = (7/99)+(7/99) = (14/99)
PS : There is no need to compute the probability of the second sequence seperately, as there are only two ways in which desired outcome can be achieved the total probability can be simply computed by multiplying the first probability by 2.
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Let me sample space S be the set of all choices of 4 balls one by one. Then |S| = (12×11×10×9)
The desired kind of alternate choices can either start with red or white : (i) The first one can arise in C(5,1) ways Thereafter a white can appear in C(7,1) ways. The next red can appear in C(4,1) ways and the next white can appear in C(6,1) ways. Hence RWRW can happen in 5×7×4×6 =840. (ii) Similarly WRWR can happen in 7×5×6×4 = 840 again. As these are disjoint possibilities, the event space E has 2×840 = 1680 elements. Therefore the required probability = |E|/|S| = 1680/12×11×10×9 = 168/12×11×9 = 14
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Let us take R for red ball and W for white ball.
Four balls can be taken one by one without replacing in the following ways.
Either WRWR or RWRW
So the required probability =
(7/12)×(5/11)×(6/10)×(4/9) +(5/12)(7/11)(4/10)(6/9)
= 7/99 + 7/99 =14/99
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