Carl has a piece of toast that has jelly on one side, and he dropped it twice. Both times, it landed with the jelly side up. If he drops it two more times, what is the probability that it will have landed jelly side up a total of three times?

##.25##

The amount of ways the toast can land jelly side up 3 out of 4 times using the binomial distribution we can look at this as the total number of success from 4 trials with probability ##p=1/2## thus

##(“_k^n)p^k(1-p)^(n-k)##

##(“_3^4)(1/2)^3(1-1/2)^(4-3)##

##4(1/2)^4 = 1/4 = .25##

You can also go through the combinations for this and see that the total number of combinations of front to back is ##2^4## which is the denominator and then the amount of ways you can get 3 out of 4 is ##(“_3^4) ## is the numerator

##((“_3^4))/2^4 =.25 ##

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Carl has a piece of toast that has jelly on one side, and he dropped it twice. Both times, it landed with the jelly side up. If he drops it two more times, what is the probability that it will have landed jelly side up a total of three times?

##.25##

The amount of ways the toast can land jelly side up 3 out of 4 times using the binomial distribution we can look at this as the total number of success from 4 trials with probability ##p=1/2## thus

##(“_k^n)p^k(1-p)^(n-k)##

##(“_3^4)(1/2)^3(1-1/2)^(4-3)##

##4(1/2)^4 = 1/4 = .25##

You can also go through the combinations for this and see that the total number of combinations of front to back is ##2^4## which is the denominator and then the amount of ways you can get 3 out of 4 is ##(“_3^4) ## is the numerator

##((“_3^4))/2^4 =.25 ##

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Your email address will not be published. Required fields are marked *

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