60.4k views
2 votes
Sixty nine percent of U.S households play video or computer games. Chose 4 heads of households at random. Find the probability that none play video/computer games?

User Coty
by
4.1k points

2 Answers

5 votes

Answer: 1*(69/100)^0*(31/100)^4 = 0.0092 or 0.0092*100 = 0.92%

User Crystyxn
by
5.9k points
4 votes

Answer: 0.00923521

Explanation:

Given : The probability U.S households play video or computer games=69%=0.69

here, the probability of each U.S household play video or computer games is fixed as 0.69

Then, the probability of each U.S household not play video or computer games= 1-0.69=0.31

For independent events the probability of their intersection is product of probability of each event.

Now, the probability that none play video/computer games will be :-


(0.31)^4=0.00923521

User Jooyoun
by
5.3k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.