sanju09 wrote:A, B, and C are solving a question. If two of them have the same chance of solving the question correctly in time, and probability that the question is finally NOT solved correctly in time is 0.252, then who of the three had a different chance of solving the question correctly in time?
(1) Probability that B will solve the question correctly in time is 0.3.
(2) Probability that only one of A, B, and C will solve the question correctly in time is 0.42.
This looks like a job for our favourite DS superhero: "Number of equations vs Number of Unknowns Man!"
Let's start with
Step 1 of the Kaplan method for DS, Analyzing the Stem.
We have a probability question, so we should jot down the probability formula:
Prob = (# of desired outcomes)/(total # of possibilities)
We have multiple things going on, so we should remind ourselves that to calculate the probability of MULTIPLE events occurring, we MULTIPLY the individual probabilities.
Applying that to the info in the stem, we know that:
.252 = Prob(A wrong) * Prob(B wrong) * Prob(C wrong)
1 equation, 3 unknowns. What would allow us to solve everything? 2 more distinct, linear equations.
(As an aside, this original equation is NOT linear - whenever you have the product of two or more variables, your equation is non-linear; however, since all probabilities are non-negative, and since in this case we also know that they're non-zero, we can safely pretend that it is linear.)
Further, we know that two of the 3 items on the right side of that equation are equal and are all in the range of:
0 < x <= 1.
Our task? To determine which one is different from the other two (or, conversely, to find out which two are the same).
On to
Step 2 of the Kaplan method for DS: Evaluate the Statements.
(2) Start with the second statement because it's the easier one to evaluate. (2) gives us general info about all three people, but no way to differentiate among A, B and C: insufficient. Hey, at least if we get stuck, we can confidently eliminate choices B and D.
(1) Now we have some info about a specific person, so we should at least stop and consider whether (1) is sufficient alone.
We know that Prob(B right) = .3; the corollary of this info is that Prob(B wrong) = .7
So, now we have 2 equations and 3 unknowns. Usually, that means insufficiency, but sometimes you can do tricky things to solve, so let's check to be sure.
If B is the "loner", then our equation becomes:
.252 = (.7)(y)(y);
does that have a solution for y inside our range of possible values? Yes, so B could be the loner.
If B is part of the matched pair, then our equation becomes:
.252 = (.7)(.7)(x);
does that have a solution for x inside our range of possible values? Yes, so B could be part of the matched pair and either A or C could be the loner.
Accordingly, (1) is insufficient alone: eliminate choice (A).
Our final step: combine the statements.
Taking the original and (1) together, we have:
.252 = (A)(C)(.7)
.252 = .7AC
We can certainly turn (2) into an equation (although a complicated one, since we'd have to sum the 3 different ways that exactly one of them could be correct). We now have 3 equations and 3 unknowns: sufficient, choose choice C.
Let's actually do at least some of the math, to demonstrate why we really do NOT want to do so on test day:
Prob(exactly 1 right) = Prob(A right)*Prob(B wrong)*Prob(C wrong) + Prob(A wrong)*Prob(B right)*Prob(C wrong) + Prob(A wrong)*Prob(B wrong)*Prob(C right)
Now, we can do a lot of substitution to clear that up.
We know that the left side is .3 and we know that Prob(B wrong) is .7 and Prob(B right) is .3.
We also know that Prob(A right) = 1 - Prob(A wrong); the same goes for C. So, letting:
Prob(A wrong) = A and Prob(C wrong) = C, we get:
.3 = (1-A)(.7)(C) + (A)(.3)(C) + (A)(.7)(1-C)
and
.3 = .7C - .7AC + .3AC + .7A - .7AC
.3 = .7A + .7C - .11AC
From our
.252 = .7AC
equation, we can turn all the AC terms into numbers, leaving us with:
(some number) = .7A + .7C
(some number)/.7 = A + C
We can now isolate either A or C and plug it back into:
.252 = .7AC
to solve for the other (we'll get a quadratic, but since our terms are all positive, that's fine). One final substitution and we have the values for A, B and C.
Sooo much math, sooo little time on Test Day: all hail "Number of equations vs Number of unknowns" Man!
