Question on binomial distributions on math!?

Question

A person randomly selexts answers to all 10 questions on a multiple choice quiz. each question has 1 correct answer and 3 incorrect answers. find the probability that the indicated number of answers are correct.

the formula

P=nCrp^r(1-p)^n-r

Question
Exactly 7

#2 Find the probability that a batter with each batting average given below will get at most 3 hits in the next 5 bats.

0.250

Answer 1

So to get exactly 7 right:

if the 7 are specified, e.g. the "first seven" or something like that, then it is simply 0.25^7 times 0.75^3

But since it can be any seven of the ten, we multiply this result by how many ways you can choose seven from 10, i.e. 10C7, which is 10*9*8/3*2*1 = 120

So we get 120 times 0.25^7 times 0.75^3 = .0030899

Answer 2

In the first question, the probability of the student to have selected the right answer is 1 of 4. In the next question also 1 of 4 up to the end. But in the first and the second question, the probability of the person having chosen the right answer is (1/4) * (1/4) = 1 of 16 so in 16 possible selections, he has only 1 chance of selecting the right answer to both questions. Now for 7 questions, he will have (1/4)^7 possibilities to chose the right answer that is equal to 0.00006103515625.

I am sure you can do the next one yourself.

Answer 3

p(7 correct) = 10C7(0.25)^7(0.75)^3 = 0.0030899 or 0.003

1*1*.75^5 = 0.2373046875
5*.25*.75^4 = 0.3955078125
10*.25^2*.75^3 = 0.263671875
10*.25^3*.75^2 = 0.087890625
P(no more than 3 hits) = 0.984375

(could be done faster by finding p(4) and P(5), adding, and subtracting total form 1)

Answer 4

for the first there is a probablity of .25 of getting the correct answer for one question so for 10 it would .25 x .25 x .25 ....ten times for an answer of 0.00000095 i think ...