While designing a game involving chance, Desmond noticed that the prob

m=2,n=4

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

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

 

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. 
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