An exam consists of 8 true/false questions. Brian forgets to

Since 70% is the passing rate, .70 ( 8 questions) = 5.6 correct answers (or 6 rounded up)

We must then look for the probability of passing:
All correct answers + 7 correct answers + 6 correct answers = probability of passing.

All correct answers = \(\frac{1}{2}^8=\frac{1}{256}\)

7 correct answers = \(\frac{8C1}{1/2^8} = \frac{8}{256}\)

6 correct answers = \(\frac{8C2}{1/2^8} = \frac{28}{256}\)

\(\frac{1}{256}+\frac{8}{256}+\frac{28}{256}=\frac{37}{256}\)

Answer is B

I am not sure if the wording of the question is accurate. Anyone else feels the same way?

There are only two possibilities - Either Brian fails or he passes. So, probability that Brian passes = 1/2

Did I read the question incorrectly or is the wording ambiguous?

OK. Got it.
On a side note, probability of you dating Charlize Theron is 0 & here is the mathematical explanation for the math expert:

Assuming Charlize Theron (C.T.) will date only one person at a time
Probability(Game dating C.T.) + Probability(Bunuel dating C.T.) + ... = 1
Since Probability(Game dating C.T.) = 1
Probability(Bunuel dating C.T.) + ... = 1 - 1 = 0

An exam consists of 8 true/false questions. Brian forgets to study, so he must guess blindly on each question. If any score above 70% is a passing grade, what is the probability that Brian passes?

(A) 1/16
(B) 37/256
(C) 5/32
(D) 219/256
(E) 15/16

Could anyone please explain?

An exam consists of 8 true/false questions. Brian forgets to study, so he must guess blindly on each question. If any score above 70% is a passing grade, what is the probability that Brian passes?

(A) 1/16
(B) 37/256
(C) 5/32
(D) 219/256
(E) 15/16

Could anyone please explain?


If you have 8 T or F and Brian is going to guess then each question he has a 50% chance of getting correct.

If a passing score is 70% it means Brian needs to get 6/8=75%, 7/8=87.5%, or 8/8=100% to pass. Each is a possibility. If Brian gets a 5/8(=62.5%) or below he fails.

So first figure out the number of ways that Brian can get 6 out of 8, 7 out of 8, and 8 out of 8 questions correct.
Which is 8 choose 6, equals is 28, 8 choose 7, equals 8, and 8 choose 8, equals 1. This sums to 37.

The number of possible questions outcomes -the sum of 8 choose 8, 7 choose 8, 6 choose 8….2 choose 8, 1 choose 8, and 0 choose 8 is 256, so the chance of him passing is 37/256.

I know there is a shorter way, but this is how I did it in about 2:34

I have a very basic and silly doubt
Why the probability of getting 7 right is 8c1/(1/2^8)

I mean i understood the denominator part but why 8c1?
As it has to be 7 answers out of 8... So i thought it will be 2^7/2^8

Please help..!!

Because 7 correct out of 8 can occur in 8 ways:
YYYYYYYN (first 7 correct and 8th not).
YYYYYYNY
YYYYYNYY
YYYYNYYY
YYYNYYYY
YYNYYYYY
YNYYYYYY
NYYYYYYY

VERITAS PREP OFFICIAL SOLUTION:

First, Brian must get 70% of the 8 questions right to pass. 70% of 8 is 5.6, though, so he must get 6 questions right (there's no such thing as "partial credit" on a true/false question). This leaves us with three possible outcomes:

6 right, 2 wrong

7 right, 1 wrong

8 right, 0 wrong

We can use our knowledge of permutations with identical elements to calculate the probability here since, for our purposes, a correct answer on the first question is the same as a correct answer on the third question. Thus, there are 8!/(6!2!), or 28, ways to get 6 questions right, 8!/7!, or 8, ways to get 7 questions right, and only 1 way to get all 8 of the questions right. Brian has a total of 37 good outcomes. The number of possible outcomes, though, is 2^8, or 256, giving Brian a 37/256 chance of passing. Good luck, Brian!

Answer: B.

Let's say that you have NO IDEA how to solve this question (many students will face this predicament).

The great thing about probability questions is that, even if you don't know how to answer them, you can often eliminate some answers by using your gut instincts alone.

Brian needs to correctly guess at least 6 of the 8 questions .
How likely does that FEEL?
Well, if you feel the likelihood is less than 50%, then we can eliminate C, D and E.

Great! In 10 seconds, we're down to a 50-50 guess between A and B.

Now, guess A or B and move on knowing that you just added a lot of time to your "time bank" that you can devote to other questions.

Cheers,
Brent

Isn't the probability of passing 1 - (the probability of failing) i.e 1 - probability of getting 3 answers wrong.

I'm not able to get the correct answer by this method. Please help

That's a start.

Getting 3 answers wrong is ONE way to fail.
Getting 4 answers wrong is another way to fail, as is getting 5 answers wrong, 6 answers wrong, etc.

So, if you want to go that route, you must deal all possible ways to fail

Brian can pass if he choose more than 5 correct answers as 70% of 8 is 5.6.
Brian chooses 6 correct answers in 8!/6!2!= 28 ways
7 correct and 1 incorrect = 8!/7! = 8 ways
8 correct answers = 1 way.
Total possible ways to answer = (2)^8 =256
Probability of Brian to pass = 37/ 256
B is the answer.

Let's solve this using counting methods.

We'll first calculate the TOTAL number of possible outcomes.
This is a true/false test, so each question has 2 possible outcomes.
So the total number of outcomes for all eight questions = (2)(2)(2)(2)(2)(2)(2)(2) = 256

Now let's deal with a numerator.
5/8 = 67.5% and 6/8 = 75%
So, in order to get above 70%, Brian must answer 6 or more questions correctly. There are three different cases to consider:

Case i: Brian correctly answers 6 out of 8.
If we let C represent a correct response, and let I represent an incorrect response, then one possible outcome is CCCCCCII, where Brian correctly answers the first 6 questions, and incorrectly answers the last 2 questions
Another possible outcome is IICCCCCC, and ICCCCICC and so on.
So the question boils down to, "In how many ways can I arrange 6 C's and 2 I's?"
---------------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
------------------------

We want to arrange the letters in CCCCCCII
There are 8 letters in total
There are 6 identical C's
There are 2 identical I's
So, the total number of possible arrangements = 8!/[(6!)(2!)] = 28
So there are 28 different ways to correctly answer 6 questions out of 8

Case ii: Brian correctly answers 7 out of 8.
From here we COULD apply the Mississippi again, but a faster way is to recognize that we just need to place one I in one of the 8 possible locations (first question or second question or third question or... etc)
So there are 8 different ways to correctly answer 7 questions out of 8

Case iii: Brian correctly answers 8 out of 8.
There's only 1 way to correctly answer 8 questions out of 8 (CCCCCCCC)

So the TOTAL number of ways to score better than 70% = 28 + 8 + 1 = 37

P(score higher than 70%) = 37/256

Answer: B

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