The probability that a voting-age adult in 2004 voted in the presidential election was 0.57. Five voting-age adults in 2004 were randomly selected. Find the probability that exactly 2 or the 5 adults voted in the presidential election.
prob(voting) = .57
prob(not voting) = .43
prob(2 of 5 voted)
= C(5,2) (.57)^2 (.43)^3
= .2583
p=probability of voting
q=(1-p)=probability of not voting.
Out of 5 adults randomly selected, the probability that exactly 2 voted is calculated according to the binomial expansion,
C(5,2)p^2q^3
=(5!/(2!3!))*0.57^2*0.43^3
=0.258 (approx.)
To find the probability that exactly 2 out of 5 adults voted in the presidential election, we can use the binomial probability formula.
The binomial probability formula is given by:
P(x) = (nCx) * p^x * (1 - p)^(n - x)
Where:
- P(x) represents the probability of getting exactly x successes
- n is the total number of trials
- x is the number of successes
- p is the probability of success on a single trial
In this case:
- n = 5 (since 5 voting-age adults were randomly selected)
- x = 2 (we want to find the probability of exactly 2 adults voting)
- p = 0.57 (probability that a voting-age adult voted in the presidential election)
To calculate the probability that exactly 2 out of 5 adults voted:
P(2) = (5C2) * (0.57)^2 * (1 - 0.57)^(5 - 2)
Let's calculate it step by step:
Step 1: Calculate the combination term (5C2).
5C2 = 5! / (2! * (5-2)!)
= 5! / (2! * 3!)
= (5 * 4) / (2 * 1)
= 10
Step 2: Substitute the values into the formula.
P(2) = 10 * (0.57)^2 * (1 - 0.57)^(5 - 2)
= 10 * 0.3249 * 0.40841
≈ 0.133378
Therefore, the probability that exactly 2 out of 5 adults voted in the presidential election is approximately 0.133378, or about 13.34%.