1.Traffic accidents occur either in the morning, afternoon, evening, or nighttim

Question

1.Traffic accidents occur either in the morning, afternoon, evening, or nighttime. The probability that an accident is in the morning is .15. The probability that an accident is in the

A. Not enough info

B..2

C..45

d..25

e. .5

2. Traffic accidents occur either in the morning, afternoon, evening, or nighttime. The probability that an accident is in the morning is .15. The probability that an accident is in the afternoon is .10. The probability that an accident is in the evening is .20. Assuming accidents are independent, what is the probability that 3 randomly selected accidents all happened in the evening?

a.3*.2=.6

b.1-.45=.55

c.2*.2*.2=.008

d..5+.2-.4=.3

e. .20

3.The probability that an individual has no illnesses is .6. What is the probability that an individual has at least one illness.?

a. Not enough info to tell

b. 1

c. .6

d. .36

e. .4

4.The probability that I buy a pint of chocolate ice cream is .30. The probability that I buy a pint of chocolate peanut butter ice cream is .60. The probability that I buy a pint of vanilla ice cream is .10. Assume that each pint I buy is independent. If I buy two pints of ice cream, what is the probability that they are the same flavor?

a. 2*(.3+.6+.1)=2

b. .3*.6*.1=.018

c. 2*(.3*.6*.1)=.036

d. .3*.3+.6*.6+.1*.1=.46

5. The probability that I buy a pint of chocolate ice cream is .30. The probability that I buy a pint of chocolate peanut butter ice cream is .60. The probability that I buy a pint of vanilla ice cream is .10. Assume that each pint I buy is independent. If I buy 4 pints of ice cream, what is the probability that at least one is chocolate peanut butter?

a. (.4)^4 = .0256

b. 1-(.4)^4 = .9744

c. (.6)^4 = .1296

d. 1-(.6)^4 = .8704

Explanation / Answer

Ans:

2)Probability of an accident in evening=0.2

Probabiity that all three happened in evening=0.2^3=0.008

Option c is correct.

3)P(atleast 1)=1-P(none)

P(atleast one illness)=1-P(no illness)=1-0.6=0.4

Option e is correct.

4)P(two are of same flavor)=0.3*0.3+0.6*0.6+0.1*0.1

=0.09+0.36+0.01=0.46

Option d is correct.

5)Probability of being chocolate peanut butter=0.6

P(atleast one is chocolate peanut butter)=1-P(none is chocolate butter)

=1-(1-0.6)^4

=1-0.4^4

=1-0.0256=0.9744

option b is correct.

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