A student wants to guess and answer six objective type questions. Each question has four options, of which, only one answer is correct. What is the probability that a student would get atleast five correct answers?
Answers
Answer:
he question asks about a probability and one of the simplest rules in probabilities is counting the favorable cases and the total number of cases.
A test with 10 questions with 4 answers each, would have a correct solution like this:
CABDCCBDAA
Also, any guess would have that form: a sequence of 10,
where each is from A/B/C/D
The total number of guesses is: 4^10
4^10 = (2^2)^10 = 2^(2*10) = 2^20 = (2^10)^2 = 1024^2
That’s about one million.
The total number of cases: about 1 million (or 1 million and something).
Now, we have to compute the number of favorable cases.
Favorable cases have 8 answers right and 2 wrong.
One doubt: The question says:
“What is the probability that he gets 8 questions correctly?”
Does it mean :
“exactly 8 correct answers to questions”
or
“at least 8 correct answers to questions” ?
This is one of the reasons I said in the beginning you have to understand very precisely what the question in the problem is…
Anyway… having “exactly 8” or “at least 8” will lead to similar results…
How many “exactly 10” cases/guesses are there? Just one!! → 1
How many “exactly 9” cases/guesses are there?
You can have the first question wrong (3 cases), the second question wrong (3 cases) … So, there are 3*10 = 30 cases → 30
How many “exactly 8” cases/guesses are there?
You can have the first question wrong (3 cases) and the second question wrong (3 cases) … this is 9 cases. The first and the third (9 cases) … with the first and other (2nd to 10th) wrong there are 9*9 = 81 cases.
With 2nd and other (not 1st: from 3 to 10… that is 8) there are 8*9 = 72 cases
Then: 7*9 … 6*9… 5*9 … 4*9….
In total:
(9+8+7+6+5+4+3+2+1)*9 = 9 * [ 9 * 10 /2] = 81*5 = 405
Another way to compute the “exactly 8” cases:
Number of different pairs of wrong answers: “from 10 choose 2” (choose 2 wrong answers from a total of 10) = 10C2 = 10!/ (8! * 2) = 10*9/2 = 45
For every pair you have 3*3 different wrong choices.
So, the total “exactly 8” is 45*9 = 450–45 = 405
And now, the results:
P(“exactly 8”) = 405 / “1 million and something” = 4/10 000 = 0.0004
P(“at least 8”) = 436 / “1 million and something” = 0.00043
You can compute the exact results with a calculator
P(“exactly 8”) = 405 / (1024*1024) = 0.000386238
P(“at least 8”) = 436 / (1024*1024) = 0.000415802
I’ll assume you mean the student gets exactly 8 questions correct. Since he is guessing, he has 1/4 chance of guessing correctly on each question and 3/4 chance of guessing incorrectly. The probability that he gets a specific 8 questions correct (and 2 incorrect) is (1/4)^8*(3/4)^2 = 9/(4^10) = 9/1048576 ~ 0.00000858306. We’re almost done!
Now, all we need is the number of ways there are to choose a specific 8 questions out of the 10 for him to answer correctly (or equivalently, to choose the 2 he will answer incorrectly). There are 10 ways to choose the first question he answers incorrectly, a
For each question we have 4 choices. So if you have 10 on such questions, you ll have :
4*4*4*4*4*4*4*4*4*4= 4^10 = 2048
Getting 8 correct means=> 8 corrects and two wrongs
3/4 is the probability if a question is answered wrong
1/4 is the probability if a question is answered correctly
so we can consider this probability :
P=(3/4)*(3/4)*(1/4)*(1/4)*(1/4)*(1/4)*(1/4)*(1/4)*(1/4)*(1/4)
=> P=0.00000858306
But as we can select any combination of 2 from 10 for being wrong, so we need to multiply P by (10,2)=10!/(2!*8!) =45
so the answer is P*45 = 0.00038623809
Step-by-step explanation:
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