Answer:
A. 0.1035
B. 0.1406
C. 0.1025
Step-by-step explanation:
Given that:
the number of sample questions (n) = 5
The probability of choosing the correct choice (p) = 1/4 = 0.25
Suppose X represents the number of question that are guessed correctly.
Then, the required probability that she gets the majority of her question correctly is:
P(X>2) = P(X=3) + P(X =4) + P(X = 5)
[tex]P(X>2) = [ (^{5}C_{3}) \times (0.25)^3 (1-0.25)^{5-3} + (^{5}C_{4}) \times (0.25)^4 (1-0.25)^{5-4} + (^{5}C_{5}) \times (0.25)^5 (1-0.25)^{5-5}[/tex]
[tex]P(X>2) = \Bigg [ \dfrac{5!}{3!(5-3)!} \times (0.25)^3 (1-0.25)^{2} + \dfrac{5!}{4!(5-4)!} \times (0.25)^4 (1-0.25)^{1} +\dfrac{5!}{5!(5-5)!} \times (0.25)^5 (1-0.25)^{0} \Bigg ][/tex]
P(X>2) = [ 0.0879 + 0.0146 + 0.001 ]
P(X>2) = 0.1035
B.
Recall that
n = 5 and p = 0.25
The probability that the first Q. she gets right is the third question can be computed as:
[tex]P(X=x) = 0.25 ( 1- 025) ^{x-1}[/tex]
Since, x = 3
[tex]P(X = 3) = 0.25 ( 1- 0.25 ) ^{3-1}[/tex]
[tex]P(X =3) = 0.25 (0.75)^{3-1}[/tex]
[tex]P(X =3) = 0.25 (0.75)^{2}[/tex]
P(X=3) = 0.1406
C.
The probability she gets exactly 3 or exactly 4 questions right is as follows:
P(X. 3 or 4) = P(X =3) + P(X =4)
[tex]P(X=3 \ or \ 4) = [ (^{5}C_{3}) \times (0.25)^3 (1-0.25)^{5-3} + (^{5}C_{4}) \times (0.25)^4 (1-0.25)^{5-4}][/tex]
[tex]P(X=3 \ or \ 4) = \Bigg [ \dfrac{5!}{3!(5-3)!} \times (0.25)^3 (1-0.25)^{2} + \dfrac{5!}{4!(5-4)!} \times (0.25)^4 (1-0.25)^{1} \Bigg ][/tex]
P(X = 3 or 4) = [ 0.0879 + 0.0146 ]
P(X=3 or 4) = 0.1025
