Math

QuestionWhat is the probability that in a random selection of 4 Americans, all, none, or at least one supports school funding? Use p=0.6p = 0.6.

Studdy Solution

STEP 1

Assumptions1. The percentage of Americans who support increased funding for public schools is60%. . We are choosing4 Americans at random.
3. Each American's opinion is independent of the others.

STEP 2

We need to find the probability that all4 of them support increased funding for public schools. This is the same as the probability that one supports it, raised to the power of4.
(All4 support)=(1 supports)4(\text{All4 support}) =(\text{1 supports})^4

STEP 3

Now, plug in the given value for the probability that one supports increased funding for public schools.
(\text{All support}) =0.60^

STEP 4

Calculate the probability that all4 support increased funding for public schools.
(All4 support)=0.604=0.1296(\text{All4 support}) =0.60^4 =0.1296

STEP 5

We need to find the probability that none of the4 support increased funding for public schools. This is the same as the probability that one does not support it, raised to the power of4.
(None support)=(1 does not support)4(\text{None support}) =(\text{1 does not support})^4

STEP 6

Now, plug in the given value for the probability that one does not support increased funding for public schools.
(None support)=0.404(\text{None support}) =0.40^4

STEP 7

Calculate the probability that none of the4 support increased funding for public schools.
(None support)=0.404=0.0256(\text{None support}) =0.40^4 =0.0256

STEP 8

We need to find the probability that at least one of the4 support increased funding for public schools. This is the same as1 minus the probability that none support it.
(At least1 supports)=1(None support)(\text{At least1 supports}) =1 -(\text{None support})

STEP 9

Now, plug in the calculated value for the probability that none support increased funding for public schools.
(At least supports)=.0256(\text{At least supports}) = -.0256

STEP 10

Calculate the probability that at least one of the4 support increased funding for public schools.
(At least supports)=0.0256=0.9744(\text{At least supports}) = -0.0256 =0.9744So, the probabilities area) All4 of them support increased funding for public schools is0.130 (rounded to three decimal places). b) None of the4 support increased funding for public schools is0.026 (rounded to three decimal places). c) At least one of the4 support increased funding for public schools is0.974 (rounded to three decimal places).

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