We can solve this problem using the binomial probability
equation:
P = [n! / (n – r)! r!] p^r q^(n – r)
where the variables are:
n = total number of questions = 10
r = number of correct = at least 5
p = probability of success = 0.5
q = probability of failure = 0.5
So what we have to do is to calculate for P for r = 5 to
10
when r = 5
P = [10! / (10 – 5)!
5!] 0.5^5 0.5^(10 – 5)
P = 0.246
when r = 6
P = [10! / (10 – 6)!
6!] 0.5^6 0.5^(10 – 6)
P = 0.205
when r = 7
P = [10! / (10 – 7)!
7!] 0.5^7 0.5^(10 – 7)
P = 0.117
when r = 8
P = [10! / (10 – 8)!
8!] 0.5^8 0.5^(10 – 8)
P = 0.044
when r = 9
P = [10! / (10 – 9)!
9!] 0.5^9 0.5^(10 – 9)
P = 9.766 x 10^-3
when r = 10
P = [10! / (10 – 10)!
10!] 0.5^10 0.5^(10 – 10)
P = 9.766 x 10^-4
So the probability that her score will be at least 5 is:
P (r≥5) = 0.246 + 0.205 + 0.117 + 0.044 + 9.766 x 10^-3 +
9.766 x 10^-4
P (r≥5) = 0.623
So about 62.3% chance.
You can do the same for the other item quiz, just set the value of n
