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Help with a probability word problem
- Thread starter strasc23
- Start date
Lets look at the same problem with two true-false questions.
A student may have the following answers: TT, TF, FT, FF.
Only one of the four results in "both correct." So the probability is 1/4.
Without listing the values, you can see there are 2 possible answers for the first question, and two for the second. So there are 2*2=4 total outcomes. Try it with 5 questions now.
A student may have the following answers: TT, TF, FT, FF.
Only one of the four results in "both correct." So the probability is 1/4.
Without listing the values, you can see there are 2 possible answers for the first question, and two for the second. So there are 2*2=4 total outcomes. Try it with 5 questions now.
HallsofIvy
Elite Member
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- Jan 27, 2012
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The probability of guessing correctly on a "T", "F" question is 1/2. To get all 5 problems correct, you have to do that on all 5 problems.
As daon2 suggests, the probability of "A and B", when A and B are independent, as here, is prob(A)*prob(B). Extend that to "A and B and C and D and E".
As daon2 suggests, the probability of "A and B", when A and B are independent, as here, is prob(A)*prob(B). Extend that to "A and B and C and D and E".
For each question P. of selecting T =1/2 So to select T each time P. =(1/2)^5=!/32A Student is taking a 5 question T/F test. If the student chooses answers at random, what is the probability of getting all questions correct? I'd like to know the formula to solve. Thanks
HallsofIvy
Elite Member
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- Jan 27, 2012
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- 7,763
In what sense did daon2 "abdicate"? Did he make an abdication speech while I wasn't looking?