Draw a circle around the ratio that is NOT EQUIVALENT to 6 to 10
1 answer:
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The first digit is one of {1, 3, 5, 7, 9}, and the last digit is one of {1, 2, 3, 4, 5}.
If the first digit is one of {1, 3, 5} (3 choices), then the last digit is one of {1, 2, 3, 4, 5} minus whatever is picked for the first digit (4 choices, and the remaining eight digits can be arranged in 8! ways, so there are 12*8! possible permutations.
If the first digit is one of {7, 9} (2 choices, then the last digit is one of {1, 2, 3, 4, 5} (5 choices), and again there are 8! ways of arranging the remaining digits, so there are 10*8! possible permutations.
Then the total number of permutations that fit the criteria is 12*8! + 10*8! = 22*8!. There are 10! total permutations that can be made overall, so the probability of randomly picking one we want is
Using conditional probability, it is found that there is a 0.1 = 10% probability that the chosen coin was the fair coin.
Conditional Probability
In which
- P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
- P(A) is the probability of A happening.
In this problem:
- Event A: Three heads.
- Event B: Fair coin.
The probability associated with 3 heads are:
out of 0.5(fair coin).
- 1 out of 0.5(biased).
Hence:
The probability of 3 heads and the fair coin is:
Then, the conditional probability is:
0.1 = 10% probability that the chosen coin was the fair coin.
A similar problem is given at brainly.com/question/14398287