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
probabiltyi think (1 Viewer)
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BlackJack
Vertigo!
Binomial theorem.
7C2 * 3^5... I think.
7C2 * 3^5... I think.
Dumbarse
Member
7C2 * 3^5
why is it that, explain
why is it that, explain
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
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
Dumbarse
Member
ohh i'm confused, ive done that question before but i did it just probability not binomial
Dumbarse
Member
i get it now but
Dumbarse
Member
yeh definately did it probablity way, forget how but
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
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
Dumbarse
Member
yehha thats it , i remember now
:giddy:
:giddy:
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.
= 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.