**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%
**