math
posted by Andrew .
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.

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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