probabiltyi think (1 Viewer)

1234567

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a test consists of 7 multiple choice questions all of which are to be attempted. each questio ncotnains 4 alternative answers of which one and only one is correct. find the number of ways in which 7 questions can be atempted so that exactly 2 questions are answered correctly
 

1234567

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see that's what i dunt get ...
coz usually isn't it
...............
OOOOOOOOOOOOOOO
HAHA,,,,i got it...our school never did that, we always used it to find probabillity...
got it......
thanks!!!!!!!!!!!!
 

1234567

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using binomial
4 choices, 1 choice is right, 3 choice is wrong

exactly 2 multiple choices right, so
that means u choose 2 right from 7 questions
and there are
1^2 to ge ttehs right
3^5 to get them wrong

using binomial,
nCk(first item)^k (second item)^(n-k)
so , the answer becomes

7C2 (1)^2 (3)^5
which is =
7c2 3^5
 

1234567

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yeah...our school hasn't done anything like that using binomial as well............
now sort of get it........
 

blitz8282

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this might be the probability way...

Ok so there are seven questions and 2 are answered correctly, so these two can be chosen in 7C2 ways.
Now the other 5 questions have to be answered incorrectly, and there are three ways to do this for any one question.
So 3.3.3.3.3 = 3^5 ways to answer these 5 questions incorrectly.

7C2 x 3^5 = 5103
 

wogboy

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The probability of getting 2/7 correct of the multiple choice quiz (with 4 alternate answers)
= 7c2 * (1/4)^2 * (3/4)^5
= 7c2 *3^5 / 4^7

[Now probability of getting 2/7 correct] = [no of possible ways of getting 2/7 correct] / [no of total ways of answering the quiz]

therefore,

[no of possible ways of getting 2/7 correct] = [probability of getting 2/7 correct] * [ no of total ways of answering the quiz]

since there's 4^7 ways of answering the quiz,

no of possible ways of getting 2/7 correct
= (7c2 * 3^5 / 4^7) * (4^7)
= 7c2 * 3^5

That's pretty much how it works.
:)
 

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