a bag contains 6 balls two are drawn and found them to be red the probability that 5 balls iñ the bag are red
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P(all balls are white)=P(remaining 2 balls are white).
No data is provided on what other coloured balls are present, or how many balls are of what colour, or whether the remaining 2 balls are to be drawn 1 at a time (2 draws) or in both in 1 single draw.
Case I:- Both balls are drawn in 1 single draw.
Then, the only possible outcomes are either: (i) both balls are white, or (ii) both balls are not white (one is white and the other is not)
So, P(both balls are white)=1/2
Thus, required probability that all balls are white is 0.5
Case II:-The 2 balls are drawn in separate draws.
Then, the only sequences of draws that are possible are: {white, non-white}, {white, white} and {non-white, white}.
But, since the question is not concerned with the order of the draws, but only with the end result, therefore {white, non-white} and {non-white, white} are one and the same.
Thus, the required probability is again 1/2.
No data is provided on what other coloured balls are present, or how many balls are of what colour, or whether the remaining 2 balls are to be drawn 1 at a time (2 draws) or in both in 1 single draw.
Case I:- Both balls are drawn in 1 single draw.
Then, the only possible outcomes are either: (i) both balls are white, or (ii) both balls are not white (one is white and the other is not)
So, P(both balls are white)=1/2
Thus, required probability that all balls are white is 0.5
Case II:-The 2 balls are drawn in separate draws.
Then, the only sequences of draws that are possible are: {white, non-white}, {white, white} and {non-white, white}.
But, since the question is not concerned with the order of the draws, but only with the end result, therefore {white, non-white} and {non-white, white} are one and the same.
Thus, the required probability is again 1/2.
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