tere are 5 white, 7 red, 4 black and 2 blue identical balls in a box . one ball is selected at random from the box . find the probability that the selected balls is (1) either white pr blue, (2) either red or black, (3) not white, (4) nither white or black.
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Answer:
tere are 5 white, 7 red, 4 black and 2 blue identical balls in a box . one ball is selected at random from the box . find the probability that the selected balls is (1) either white pr blue, (2) either red or black, (3) not white, (4) nither white or black.
What is the probability of 2 are red and one is blue
Work it out first where the order they are drawn is important.
RRB=718×617×316RRB=718×617×316
Then work out possible different orders they could be drawn RRB, RBR, BRR, there are three permutations P(3,1). so the answer is 7272×3=21272=0.07727272×3=21272=0.0772 = 7.72%
At least 2 are white
This is the probability that 3 are white plus the probability that 2 are white.
The probability that 3 are white is:
818×717×616=7102818×717×616=7102
The probability that exactly 2 are white is:
818×717×1016×C(3,2)=35306×3=35102818×717×1016×C(3,2)=35306×3=35102
Answer: 1. + 2. = 7102+35102=42102=717