A statistics professor gives a surprising quiz consisting of 10 true/false questions and states that passing r

Question

A statistics professor gives a surprising quiz consisting of 10 true/false questions and states that passing requires 6 correct responses. Assume that an unprepared student adopts the questionable strategy of guessing for each answer.
Find the probability that the first 6 responses are correct and the last 4 are wrong.

Choose the correct statement

a. The probability of guessing correctly on the first 6 questions and incorrectly on the remainder is less than the probability of passing by guessing.

b. The probability of guessing correctly on the first 6 questions and incorrectly on the remainder is greater than the probability of passing by guessing.

c. The probability of guessing correctly on the first 6 questions and incorrectly on the remainder is the same as the probability of passing by guessing.

Answer 1

As others have said, the probability that the first 6 are correct and the last 4 are wrong is 1/2^10 = 1/1024.

This is much smaller than the probability of passing by guessing, because there are many other ways in which you can get 6 correct questions and 4 incorrect ones; and you could also get 7, 8, 9 or 10 questions correct by guessing. Therefore (a) is correct and (b) and (c) are wrong.

Answer 2

1/(2)^10 = 1/1024
as I understand the question

Answer 3

The chance that out of:
(T/F)(T/F)(T/F)(T/F)(T/F)(T/F)(T/F)(T/F)(T/F)(T/F)
You get
(T)(T)(T)(T)(T)(T)(F)(F)(F)(F)
Guessing on each problem is merely:
(1/2)^10
or .0009765624
Essentially, there are 2^10 possible results, and TTTTTTFFFF is one of them. Another way to thing of this is 1/(2^10) which is also .0009765624.

Answer 4

c

Answer 5

Id like to use my lifeline phone-a-friend.......wtf is this Q about again???????????