Problem Solving

11 fair coins are tossed together, find the probability that there are more heads than tails?

Dont have answer choices. Can someone explain?

shankar.ashwin wrote:11 fair coins are tossed together, find the probability that there are more heads than tails?

Dont have answer choices. Can someone explain?

There is a total of 2^11 different combinations of 11 tosses. Think of it as 2^11 different "strings" of H and T:

TTTTTTTTTTT
TTTTTTTTTTH
TTTTTTTTTHH
TTTTTTTTHHH
THTHTHTHTHT

etc.
etc.

In some of them there are more Ts then Hs, and in some of them there are more Hs than Ts, but with 11 tosses (an odd number) there cannot be a tie - you cannot have an equal number of Hs and Ts. If these are the only two options: T>S or T<S, and the coins are fair, then there REALLY IS NO REASON WHY ONE SCENARIO WILL BE MORE FAVORABLE THAN THE OTHER. The probability of each scenario is equal to the other, and both are equal to half.

You can always verify what Geva said by actually solving the problem.

The outcome that we need is atleast 6 heads. It will be all of the following scenarios
Since this is a binomial probability, the probability of any outcome is 2^11 irrespective of whether we get heads or tails

6 Heads and 5 Tails
HHHHHH TTTTT Which can be arranged in 11!/(6!5!)= 462 Ways. And the final probability of this outcome is 462/(2^11)

7 Heads and 4 Tails
HHHHHHH TTTT Which can be arranged in 11!/(7!4!)=330 Ways. And the final probability of this outcome is 330/(2^11)

8 Heads and 3 Tails
HHHHHHHH TTT Which can arranged in 11!(8!3!)=165 Ways. And the final probability of this outcome is 165/(2^11)

9 Heads and 2 Tails
HHHHHHHHH TT Which can be arranged in 11!/(9!2!)=55 Ways. And the final probability of this outcome is 55/(2^11)

10 Heads and 1 Tail
HHHHHHHHH T Which can be arranged in 11!/10!=11 Ways. And the final probability of this outcome is 11/(2^11)

11 Heads
HHHHHHHHHHH Which can be arranged in only 1 Ways. And the final probability of this outcome is 1/(2^11)

Adding all of the above we get (462+330+165+55+11+1)/(2^11)=1024/(2^11)=
(2^10)/(2^11)=[spoiler]1/2[/spoiler]

Okay if the same sum was asked for 10 coins tossed,

Excluding cases of getting exactly 5 heads and 5 tails we have 10! /(5! * 5!) = 252 cases.

Therefore remaining cases = 2^10 - 252 = 772.

Out of these 772, exactly 386 of the cases would have greater number of heads and the other 386 cases would have more tails.

Therefore, P(More No of heads) would be 386/1024 = 193/512.

Is this correct?

@Ashwin
Brother, ur method is perfect. But honestly, that logical approach that Geva used works well when u clearly know the underlying principals in and out. First use a pen and paper and solve and once u master the fundas, then go to use the logical approaches. That's my opinion at atleast. Since u have the time while u are practicing, u might as well clear up all the doubts now.

For 10 coin tosses
Since it is a binomial probability, total possible outcomes are 2*2*2*2*2*2*2*2*2*2=1024

We need more heads than tails. So we need all of the following scenarios

6H 4T
HHHHHH TTTT = 210 ways

7H 3T
HHHHHHH TTT = 120 ways

8H 2T
HHHHHHHH TT = 45 Ways

9H 1T
HHHHHHHHH T = 10 Ways

10 H
HHHHHHHHHH = 1 Way

Total=386/1024 = 193/512