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%
