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(1 point) Assume that Tom attends class randomly with probability 0.55 and that each decision is independent of previous attendance, so that the process can be viewed as a Bernoulli process. What is the probability that he attends at least 7 of 10 classes given that he attends at least 2 but not all 10 classes?

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scanram

Answer:

The answer is ≈0.263500.99296≈0.26537≈26.54%

Step-by-step explanation:

We’ll use the binomial probability relation of Tom attending his classes.

n             k        n−k  

∑  Cn,k(p) (1−p)         =1

k=0

Range of probabilities

n=10, k= number of classes he attends

p=0.55.  

10                   k         n−k  

∑  C10,k(0.55) (0.45)      =1

k=0

the probability of his attending "all his classes" isn't 1 - we need to subtract out the classes we know he won't attend and set that as the denominator:

1−(C10,0(0.55)0(0.45)10+C10,1(0.55)1(0.45)9+C10,10(0.55)10(0.45)0)≈1−0.00704≈0.99296

Now the numerator. We're asked for the probability that he attend at least 7 of his classes, but we're also told he won't attend all 10, and so we'll sum up for 7≤k≤9

C10,7(0.55)7(0.45)3+C10,8(0.55)8(0.45)2+C10,9(0.55)9(0.45)1≈0.26350

And so the probability that he attends at least 7 classes, knowing he'll attend at least 2 but not 10 is: ≈0.263500.99296≈0.26537≈26.54%

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