Posted by **Matt** on Sunday, May 13, 2012 at 6:58pm.

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.

- MATH -
**Reiny**, Sunday, May 13, 2012 at 9:35pm
prob(voting) = .57

prob(not voting) = .43

prob(2 of 5 voted)

= C(5,2) (.57)^2 (.43)^3

= .2583

- MATH -
**MathMate**, Sunday, May 13, 2012 at 9:36pm
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.)

## Answer this Question

## Related Questions

- math - The probability that a voting-age adult in 2004 voted in the presidential...
- statistics - the probability that a voting-age adult in 2008 voted in the ...
- MATH - IN LOCAL TOWNS 4/5 OF THE VOTING AGE POPULATION IS REGISTRED TO VOTE 11/...
- Math 1330 - In a survey of 1000 eligible voters selected at random, it was found...
- Statistics - 12% of voters in a recent election voted on the greens candidate, ...
- mATH - First graders who misbehave in school may be more likely to be regular ...
- statistics - First graders who misbehave in school may be more likely to be ...
- Statistics - In a survey of 1233 people, 917 people said they voted in a recent ...
- statistics - First graders who misbehave in school may be more likely to be ...
- math - Fifteen percent of the seniors voted for Joe. If 289 seniors voted for ...