99 unbiased (equal likelihood of heads and equal likelihood of tails)

Question

99 unbiased (equal likelihood of heads and equal likelihood of tails) are tossed one by one. What is the probability of getting at least 50 heads?

A.  \frac{1}{2}

B.  \frac{50}{99}

C.  \frac{1}{3}

D.  \frac{2}{9}

E. None of these

Regor60 · Answer

If you call the probability of having 50 or more heads X, then 1-X must be the probability of 49 or fewer heads.

But 49 or fewer heads is the same thing as 50 or more tails.

And if you believe 50 or more tails is just as likely as 50 or more heads then you can say that:

1-X=X and therefore X=1/2

Posted from my mobile device

Nikkz99 · Answer

Could you please explain even bit further?
 Bunuel   Krunaal   KarishmaB ?!
Thank You!

Krunaal · Answer

Let us consider an example with smaller numbers and try to understand, suppose we have 3 unbiased coins and we want to find the probability of getting atleast 2 Heads.

Total cases:  2^3 = 8  (since each coin is unbiased and can be either H or T)

Atleast 2 H: HHT, HTH, THH, HHH => 4 ways

Probability =  \frac{4}{8} = \frac{1}{2}  .......................(i)

Now what if they had asked us probability of 2 Tails? which can also be said as probability of getting less than 2 Heads

Atleast 2 T: TTH, THT, HTT, TTT => 4 ways

Probability =  \frac{4}{8} = \frac{1}{2}  ........................(ii)

Did you notice? Probability of getting atleast 2 H = Probability of getting atleast 2 T = Probability of getting less than 2 H

Probability of getting atleast 2 H + Probability of getting atleast 2 T = 1 ........................from (i) and (ii)

Hence probability of getting atleast 2 H is  \frac{1}{2}

While i don't have the conceptual knowledge of in which all cases this symmetry would follow (an expert reply might help with that), but i took smaller odd number of tosses to scale up the logic  Regor60  followed.

Nikkz99 · Answer

Thank You,  Krunaal  , for the swift respone.

KarishmaB · Answer

The answer here will be simply 1/2 because of the principle of symmetry. Here is a discussion on it: https://anaprep.com/combinatorics-linea ... -symmetry/ When I toss a coin, a Heads or a Tails is equally likely. So they are equivalent elements. When we toss a coin 99 times, there will be cases {0 H, 99 T} and {99H, 0 T} - Both equally likely {1 H, 98 T} and {98H, 1 T} - Both equally likely {2 H, 97 T} and {97H, 2 T} - Both equally likely ... {49 H, 50 T} and {50 H, 49 T} - Both equally likely They are asking us what is the probability of getting the second half of the cases - 50 or more Heads. Well it is equal to the probability of the first half of the cases and total probability of all these cases is 1. So our required probability = 1/2 Answer (A)

Nikkz99 · Answer

Thank You!! I understood now as well as came to know about <Symmetry> concept.

hr1212 · Answer

Symmetry method is faster but if anyone is interested in mathematical proof, we can do -

P = Favorable outcomes/Total outcomes

Total outcomes = 2*2*....*2 (99 times) =  2^{99}

Favorable outcomes = 99C50 + 99C51 + 99C52 + .... + 99C99 (ie. selecting 50 heads + selecting 51 heads + .... + selecting 99 heads)

Also it's good to remember => nC0 + nC1 + nC2 + .... nCn =  2^n

By which,

99C0 + 99C1 +99C2 + .... + 99C49 + 99C50 + 99C51 + .... + 99C99 =  2^{99}

which can be rewritten as,

2*(99C50 + 99C51 + ... 99C99) =  2^{99} 
99C50 + 99C51 + ... 99C99 =  2^{98}

P = \frac{2^{98}}{2^{99}} = \frac{1}{2}

IMO: A

Vitthal19 · Answer

when we do the same experiment with 4 coins and we need atleast 3 heads,
so the solution would be
(1/2)^4*4c3 + (1/2)^4*4c4
(1/2)^4*  
(1/16)*5
5/16

this is how I think we have to apply the combination formula in the question could you please tell is this right or the example I have taken and the case asked in question differs?

adconsectetur · Answer

in total there are 100 possible heads. 0H = 99T and 99H = OT.

From 0 to 99 is the sample space. =100

Favorable is from 50 to 99 = 50

Probability = 50/100

hope it helps

99 unbiased (equal likelihood of heads and equal likelihood of tails)

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99 unbiased (equal likelihood of heads and equal likelihood of tails)

Prep Toolkit

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