Posted by **Andrew** on Tuesday, November 20, 2012 at 6:48am.

Beth Dahlke is taking a ten question multiple choice test for which each question has three answer choice only one of which is correct. Beth decides on answers by rolling a fair die and making the first answer choice if the die shows 1 or 2 the second if it showes 3 or 4 and the third if it showes 5 or 6. Find the probability of each event a) exactly fseven correct answers b) at least seven correct answers.

- math -
**Reiny**, Tuesday, November 20, 2012 at 8:43am
Prob of 1 or 2 on a die = 2/6 = 1/3

prob of 2 or 4 = 1/3

prob of 5 or 6 = 1/3

Since these become your choice of answer

prob of selecting 1st answer = 1/3

prob of selecting 2nd answer = 1/3

.....

so the prob of choosing the correct answer in each event is simply 1/3

so to get 7 out of 10 correct

= C(10,7) (1/3)^7 (2/3)^3 = appr .01626

so at least 7

= C(10,7) (1/3)^7 (2/3)^3 + C(10,8) (1/3)^8 (2/3)^2 + .. + C(10,10) (1/3)^10

I will let you do the button pushing

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