While designing a game involving chance, Desmond noticed that the

Question

While designing a game involving chance, Desmond noticed that the probability a fair coin lands faceup exactly 2 times when the coin is tossed 3 times is equal to the probability that a fair coin lands faceup exactly m times when the coin is tossed n times, where n > 3.

Select for m and for n values consistent with the given information. Make only two selections, one in each column.

Select for  m  and for  n  values consistent with the given information. Make only two selections, one in each column.

ID: 700252

egmat · Accepted Answer

This is another one of those quant probability questions in the garb of TPA question. And it surely has several layers of complexity.
 
 Translation while reading and drawing inferences about the scene expressed about probability.
 
 Were you able to formulate the expressions of probability correctly?   
 Probability by itself is a fretted upon concept. And on top of that this question adds another layer of complexity with variables.
 
 Did you get scared of these variables m and b here?   
 The data set presents the variables and indicates that n > 3. This may seem like an inconsequential bit of information, but it is indeed very important. In fact, if you do not pay attention to this information, you may end up selecting incorrect answer.
 
 Did you understand the relevance of this n > 3, while you were reading the dataset?   
 The next layer of complexity comes from the conceptual arena of probability. While calculating the probability of the first scenario, one does not account for the order.
 
 Did you miss accounting for the order in which the ups and downs appear?   
 Now when you process the equation, you get 1 equation with 2 variables and that too with combination processor. This means that we need to have a solid approach of analyzing the answer choices.
 
 Did you understand how process the choices with variables in combination formula?

A total of 5 layers of complexity – all packed in a single question. Such a treasure trove of learnings!

https://www.youtube.com/watch?v=poQFXm2OYoI

KarishmaB · Answer

Probability a fair coin lands faceup exactly 2 times when the coin is tossed 3 times Total outcomes when a fair coin is tossed 3 times = 2^3 = 8 Favorable outcomes = 3!/2! = 3 (HHT or HTH or THH) So required probability = 3/8 This is a bit less than 1/2. When will we get the same probability if the coin is tosses 4 times or 5 times or 6 times? Consider a coin tossed 4 times. Since it is fair coin, the probability of getting only 1 Heads (less likely) will be significantly lower than that of getting 2 Heads (more likely) Total outcomes when a fair coin is tossed 4 times = 2^4 = 16 Favorable outcomes = 4!/2!*2! = 6 (HHTT or HTTH etc.) So required probability = 6/16 = 3/8 It matches and hence we select m = 2 and n = 4. ANSWER Note that as the number of tosses increases, this probability will keep reducing. At 5 coin tosses, the highest probability will be of gettig 2 Heads (will be the same as that of getting 3) Heads which will be 5/16 - much lower than 1/2 This is so because there are many other possibilities - Only 1 Heads (same as 4 Heads) or No Heads (same as All Heads) When there are 6 tosses, the probability of getting 3 Heads will be 5/16 again - much lower than 1/2 The logic is same as above. Another Combinatorics based TPA: https://youtu.be/WassWEtmcGo

jaky_nguyen · Answer

P of 2 UP in 3 tosses: 3C2/ 2^3= 3/8 
- if n =4>3; m=3; P of 3 UP in 4 tosses: 4C3/2^4 = 4/16=1/4 # P of 2 UP in 3 tosses
- if n =4; m=2; P of 2 UP in 4 tosses: 4C2/2^4 = 6/16=3/8 
=> m=2; n=4

Elite097 · Answer

P of exact 2 heads in 3 tosses= (1/2)^3*3!/2!= 3/8= 4/16=8/32=16/64

Now, per question n can be 4,5 or 6 since it is >3
If n=4 and m=3 faceups , P(HHHT)= (1/2)^4*4!/3!=4/16 which is not equal to 6/16
If n=4 and m=2 faceups , P(HHTT)= (1/2)^4*4!/2!*2!=6/16 which is equal to 6/16

Hence n=4, m=2

Gemmie · Answer

faceup exactly 2 times when the coin is tossed 3 times

No. of total arrangements:  2^3 = 8 
No. of exactly 2 faceup arrangements: 3C2 = 3

=> Probability:  \frac{3}{8}

=  \frac{6}{16} = \frac{6}{2^4}

=  \frac{12}{32} = \frac{12}{2^5}

=  \frac{24}{64} = \frac{24}{2^6}

If n = 4 
6 = 2*3 => Check 4C3 = 6
 => m = 3; n = 4

Adarsh_24 · Answer

I think may be we could do something like this too.

3/8 =   * (1/2^n)
m!(n-m)! = 8*n! / 3*2^n

N>3
N probably not 5
4 or 6

If n=4,
m!(4-m)! = 4
So m can be 2

Ans 4,2

But let’s assume it weren’t 
Then n=6
m!(6-m)! = 30
1*30
2*15
3*10
5*6
Don’t see any factorial combos.

agrasan · Answer

When we toss a fair coin to get exactly 2 heads, it can be HHT, HTH, THH
So P(exactly 2H) = ((1/2)^3)*(3!/2!) --------- (eq 1)

When we toss a fair coin n times to get exactly  m heads  or n-m tails
arrangements can be done in (n!)/((m!)*(n-m!))
So P(exactly m heads) = ((1/2)^m)*((1/2)^n-m)*(n!)/((m!)*(n-m!)) = ((1/2)^n)*(n!)/((m!)*(n-m!))

Further simplifying we get 
((1/2)^3)*((1/2)^n-3)*(n!)/((m!)*(n-m!)) ------ (eq 2)

If we equate eq (1) and eq (2)

((1/2)^3)*(3!/2!) = ((1/2)^3)*((1/2)^n-3)*(n!)/((m!)*(n-m!))

3 = ((1/2)^n-3)*((n!)/((m!)*(n-m!)))

We are clearly told that n>3

We can get a rough idea that we don't want 5 on RHS in numerator so n=5/6 is not possible but we do want 3 in RHS numerator
By trying n = 4, m=2 we can see LHS = RHS

While designing a game involving chance, Desmond noticed that the

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