SOLUTION: Question from Binomial Probability
If three coins are tossed, find the probability of each number of heads.
4) 3
5) 1 or 2
6) at least 1
According to the book, For number 5
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-> SOLUTION: Question from Binomial Probability
If three coins are tossed, find the probability of each number of heads.
4) 3
5) 1 or 2
6) at least 1
According to the book, For number 5
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Question 1001201: Question from Binomial Probability
If three coins are tossed, find the probability of each number of heads.
4) 3
5) 1 or 2
6) at least 1
According to the book, For number 5 the answer is 3/4.
would like to see solutions shown for all three. Thanks.
You can put this solution on YOUR website! The binomial probability formula is
P(k successes in n trials) = nCk * p^k * q^(n-k) where
n = number of trials
k = number of successes
n-k = number of failures
p = probability of success in one trial
q = 1 – p = probability of failure in one trial
nCk = combination of n trials taken k at a time
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for our problem
n = 3
p = 0.50
q = 0.50
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4) 3 successes
P(3 successes in 3 trials)= 3C3 * (0.50)^3 * (0.50)^(3-3) =
(3! / (3! * (3-3)!) * (0.125) * 1 =
0.125
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5) first calculate
P(0 successes in 3 trials) = 3C0 * (0.50)^0 * (0.50)^(3-0) =
(3! / (0! * (3-0)!) * 1 * (0.125) =
0.125
therefore
P(1 or 2 successes in 3 trials) = 1 - (P(3 successes in 3 trials) + P(0 successes in 3 trials)) =
1 - (0.125 + 0.125) = 0.75 = 3/4
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6) note that
P(at least 1 success in 3 trials) = 1 - P(0 successes in 3 trials) =
1 - 0.125 = 0.875
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