We need to find the probability that the student guesses correctly on 9 or 10 true false questions:

(1) The most straightforward way is to use the binomial distribution. Given the probability p of some event e, the probability of getting x events in n trials is given as:

`P(e=x)=...

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We need to find the probability that the student guesses correctly on 9 or 10 true false questions:

(1) The most straightforward way is to use the binomial distribution. Given the probability p of some event e, the probability of getting x events in n trials is given as:

`P(e=x)= ` `_nC_x (p^x) (1-p)^(n-x) `

Here x=9 or 10, n=10, and the probability is p=.5 as there is an equally likely chance of getting the question right or wrong. Since we have 9 or 10, and the events are mutually exclusive, the total probability is P(x=9)+P(x=10).

`P(x=9)= ` `_(10)C_9 .5^9 .5^1=(10)(1/512)(1/2)=5/512~~.00977 `

`P(x=10)= ` `_(10)C_(10)(.5)^10(.5)^0=1/1024=9.765625"x"10^(-4) `

**So the probability we seek is approximately .00977+.000977=.01074**

My calculator gives .0107421875 as the approximate probability (approximate only in that the digits keep going; the exact answer is `5/512+1/1024=11/1024 `) .

(2) A second way to approach this is to use the multiplication principle. There are 10 events.he likelihood of getting all questions correct is the product of the probabilities of getting each answer correct. Since each event is independent we get `(.5)^10=1/1024 `

The probability of getting 9 correct is a little bit more complicated. Again we take the product of the probabilities of each event. The one incorrect response also has a probability of .5 so the product is `1/1024 ` . However, there are 10 ways this could have happened: the incorrect response could have been on question 1, 2,3,...,10. So we multiply the product by 10 to get `5/512 ` as before.

(3) We could use the binomial approximation, though you would not use this in this case.

Here n=10, p=.5, 1-p=.5 so `mu=5,sigma~~1.581 ` . Then we want `P(x>=9)==>P(x>=8.5) ` using the conversion for continuity. Then `z=(8.5-5)/1.581~~2.21 ` so we find P(z>2.21) in a standard normal distribution table to be approximately .013.

We can use this approximation since np=n(1-p)>5, but typically you would compute the exact probability as in (1) and (2).

**Further Reading**