A fair coin has 2 distinct flat sidesone of which bears the image of

Question

A fair coin has 2 distinct flat sides—one of which bears the image of a face and the other of which does not—and when the coin is tossed, the probability that the coin will land faceup is ½. For certain values of M, N, p, and q, when M fair coins are tossed simultaneously, the probability is p that all M coins land faceup, and when N fair coins are tossed simultaneously, the probability is q that all N coins land faceup. Furthermore, M < N and 1/p + 1/q = 72.

In the table, select a value for M and a value for N that are jointly consistent with the given information. Make only two selections, one in each column.

OA:
M=3
N=6

In the table, select a value for  M  and a value for  N  that are jointly consistent with the given information. Make only two selections, one in each column.

ID: 100444

AdityaHongunti · Accepted Answer

Trial and error would be quicker.

Probability for each event irrespective of the coin ) M or N is 1/2

i.e. if M =1 coin 
> p (probability of face) =1/2

Similarly for N

Now for M coins 
> p= (1/2)^M

For M coins
> p= (1/2)^N

Now to confuse us , the reciprocal of sums is given so that we get confused.
But one clue is that the sum is 72, an integer, so you need not fear.

1/p +1/q =72

1/(1/2)^M + 1/(1/2)^N = 72

>> 2^M +2^N = 72.

All you have to do list down some powers of two.not much just few.

2
4
8
16
32
64

Given M<N

Trial and error gives us 
8+64 =72
2^3 + 2^6 = 72

M=3 N=6

Posted from my mobile device

egmat · Answer

This is a Quant Probability question in the garb of a TPA question. So, as long as you have solid quant skills, you should be able to solve this question with utmost ease.

Here is the video solution for you!

https://www.youtube.com/watch?v=vuZWoL_8Avc

If you falter in this question, you should implement the stated corrective actions and once you do that not only will your ability in TPA improve but also in quant Arithmetic. 😊
 
 If you were not able to translate the scenario into probability equation, then the issue is your lack of conceptual understanding related to probability.
 
 Corrective Action: Build your probability conceptual understanding.   
 If you got scared of this question because you saw 4 different variables in a probability question, then the issue is that you are not comfortable using variables in word problems type questions.
 
 Corrective Action: Build your process skills in the realm of word problems.

In addition, this question can be solved using two methods:
 
 Using the answer choices 
 Using inferences

Watch the solution to learn how these two methods can be used.

chetan2u · Answer

Probability that all M have same outcome = 1/2^m
And probability that all N have same outcome = 1/2^n
So 1/(1/2)^m + 1/(1/2)^n = 72 thus 2^m+2^n=72
2^m(1+2^{n-m})=72=2^3*3^2... So m=3
So 1+2^{n-m} must be a multiple of 9.. so 2^{n-m}=8....
n-m=3...n=6

kritisinghvi · Answer

Another way would be trial basis...

By solving the given equation, we can say p+q=72pq or q=p/(72p-1)

We also know p and q will have denominators in the form of 2^x

Start with option A)
If M=2, then
 p= (1/2)^2 = 1/4 -> q=1/88 - not possible as denominator not of the form 2^x

Similarly, Option B)
If M=3, then
 p=1/8 -> q=1/64 = 1/(2^6)
 Therefore N = 6.

Answer:
 M = 3 (Option B)
 N = 6 (Option E)

Posted from my mobile device

arpitc · Answer

This way of solving looks more straightforward.

Fish181 · Answer

I recognized that 1/p + 1/q = 72 basically says the denominators of the probability equal 72 (because 1/FRAC = the denominator).

I then also recognized that since the probability being multiplied is 1/2, the denomenators of the probabilities will only include the prime factor of two. (2 4 8 16 32 64 etc)

I started playing with the numbers and found that 8 + 64 = 72. 8 and 64 are both numbers with only 2 as their factor. 2^3 = 8 and 2^6 = 64 therefore the answers are 2 and 6. m<n therefore 2=m 6=n.

Dbrunik · Answer

This is a very high quality question that tests your ability to use algebra and probability concepts jointly. do you understand how the translate a series of isolated coin flips into an algebraic expression, and how variables effect the outcome of a certain event?

a coin flip is obviously 1/2 odds of being face up, but put with 1/p and 1/q = 72, can you ascertain that P is some variable that is influenced by m? and how does it fit together with N given the available answer choices?

great question, very well written, and is essential for understanding probabilities

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A fair coin has 2 distinct flat sidesone of which bears the image of

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