Asked by Zul
There are two independent multiple-choice quizzes where quiz 1 has eight
questions and quiz 2 has 15 questions. Each question in the first quiz has four
choices and each question in the second quiz has five choices. Suppose a student
answers the questions in the quizzes by pure guessing.
i. What is the probability that at most three questions must be answered to
obtain the first correct answer in quiz 1? Interpret its value.
ii. What is the probability that less than four correct answers in quiz 2?
iii. In order to get high score, Siti needs to obtain at least six correct answers in
quiz 1. Is Siti likely to get the high score in quiz 1?
iv. What is the probability that obtains five correct answers before eighth
questions in quiz 2? Interpret its value.
v. Based on your answer in (iv), would you consider this event likely to occur?
Explain your reason.
questions and quiz 2 has 15 questions. Each question in the first quiz has four
choices and each question in the second quiz has five choices. Suppose a student
answers the questions in the quizzes by pure guessing.
i. What is the probability that at most three questions must be answered to
obtain the first correct answer in quiz 1? Interpret its value.
ii. What is the probability that less than four correct answers in quiz 2?
iii. In order to get high score, Siti needs to obtain at least six correct answers in
quiz 1. Is Siti likely to get the high score in quiz 1?
iv. What is the probability that obtains five correct answers before eighth
questions in quiz 2? Interpret its value.
v. Based on your answer in (iv), would you consider this event likely to occur?
Explain your reason.
Answers
Answered by
Reiny
for test1
prob correct = 1/4
prob wrong = 3/4
i)
could be
c or wc or wcc , (c for correct, w for wrong)
prof (of that)
= 1/4 + (3/4)(1/4) + (3/4)^2 (1/4
= 37/64
ii) less than 4 correct
---> none correct + 1 correct + 2 correct + 3 correct
= C(8,0) (1/4)^0 (3/4)^8 + C(8,1)(1/4) (3/4)^7 + C(8,2)(1/4)^2 (3/4)^6 + C(8,3)(1/4)^3 (3/4)^5
= .88618
iii) , what she CAN'T have is 7 wrong or 8 wrong.
prob of that
= 1 - C(8,7)(3/4)^7 (1/4 ) - C(8,8) (3/4)^8
= .6329
Follow the same kind of reasoning steps for the rest
prob correct = 1/4
prob wrong = 3/4
i)
could be
c or wc or wcc , (c for correct, w for wrong)
prof (of that)
= 1/4 + (3/4)(1/4) + (3/4)^2 (1/4
= 37/64
ii) less than 4 correct
---> none correct + 1 correct + 2 correct + 3 correct
= C(8,0) (1/4)^0 (3/4)^8 + C(8,1)(1/4) (3/4)^7 + C(8,2)(1/4)^2 (3/4)^6 + C(8,3)(1/4)^3 (3/4)^5
= .88618
iii) , what she CAN'T have is 7 wrong or 8 wrong.
prob of that
= 1 - C(8,7)(3/4)^7 (1/4 ) - C(8,8) (3/4)^8
= .6329
Follow the same kind of reasoning steps for the rest
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