a bag only contains black and white counters. A counter is chosen from the bag at random and replaced. Another counter is chosen from the bag at random. The probability of choosing two black counters is 0.36 Show that the probability of choosing a black counter each time is 0.6
2) Work out the probability of choosing 2 white counters.
3) Work out the probability of choosing at least one white counter?
This is just to help us revise, I've tried to work it out but i'm stuck can someone help please?
2) Work out the probability of choosing 2 white counters.
3) Work out the probability of choosing at least one white counter?
This is just to help us revise, I've tried to work it out but i'm stuck can someone help please?
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Drawing a tree diagram isn't essential but it will help you to understand this question.
1) P(BB) = 0.6 x 0.6 = 0.36
2) P(WW) = 0.4 x 0.4 = 0.16
3) P(at least one white) = P(not both black) = 1 - 0.36 = 0.64
(or you can add 0.16 + 0.24 + 0.24)
Good luck with Unit 1 test.
1) P(BB) = 0.6 x 0.6 = 0.36
2) P(WW) = 0.4 x 0.4 = 0.16
3) P(at least one white) = P(not both black) = 1 - 0.36 = 0.64
(or you can add 0.16 + 0.24 + 0.24)
Good luck with Unit 1 test.
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1. Let the probability of choosing a black counter = p.
We know that the probability of choosing a black counter is independent across each trial (since the counter is replaced), so the probability of choosing a black counter twice in a row is p squared.
We know that p^2 = .36 so p = .36^(1/2)
2. The probability of choosing a white counter is 1 - p. The probability of choosing two white counters in a row is (1-p)^2. This follows the same logic as above.
3. I think this question is asking: " If you pull out two counters independently, what is the probability that one of them is white?"
Rephrasing this, what is the probability that they are not both black?
The probability of choosing two black counters is 0.36, which is already given to us. Then the probability that one of them is white (i.e. they aren't both black) is 1 - .36
We know that the probability of choosing a black counter is independent across each trial (since the counter is replaced), so the probability of choosing a black counter twice in a row is p squared.
We know that p^2 = .36 so p = .36^(1/2)
2. The probability of choosing a white counter is 1 - p. The probability of choosing two white counters in a row is (1-p)^2. This follows the same logic as above.
3. I think this question is asking: " If you pull out two counters independently, what is the probability that one of them is white?"
Rephrasing this, what is the probability that they are not both black?
The probability of choosing two black counters is 0.36, which is already given to us. Then the probability that one of them is white (i.e. they aren't both black) is 1 - .36
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Since the probability of choosing one black counter is 0.6, the probability of choosing 2 black counters is 0.6 x 0.6 = 0.36. The probability of choosing 1 white counter is then 1 - 0.6 = 0.4, and 2 white counters is 0.4 x 0.4 = 0.16.
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1) P(BB) = 0.6 x 0.6 = 0.36
2) P(WW) = 0.4 x 0.4 = 0.16
3) P(at least one white) = P(not both black) = 1 - 0.36 = 0.64
(or you can add 0.16 + 0.24 + 0.24)
Hope that helps!
2) P(WW) = 0.4 x 0.4 = 0.16
3) P(at least one white) = P(not both black) = 1 - 0.36 = 0.64
(or you can add 0.16 + 0.24 + 0.24)
Hope that helps!