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Re: If all the coins are fair, in that the probability of heads = probab [#permalink]
1
It sure will.
Ex:
For option A, we need 11 heads from 24 flips so probability will be \(\frac{11}{24}\)
&
for option C, we need 14 tails so probability will be \(\frac{14}{24} = \frac{7}{12}\)

I hope this is what you wanted to ask. Please ask if the doubt remains.

forwet wrote:
How the probabilities would differ if all the possible outcomes have same probability?

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If all the coins are fair, in that the probability of heads = probab [#permalink]
Given that all coins are fair and probability of “heads” = probability of “tails” = \(\frac{1}{2}\) and we need to find which of the following events has the greatest probability?

Since the coins are fair so at each and every toss P(H) = P(T) = \(\frac{1}{2}\)

So, irrespective of how many tosses we do, Number of Heads = Number of Tails = \(\frac{1}{2}\) * Number of tosses

=> After 24 tosses, Number of Heads = Number of Tails = \(\frac{1}{2}\) * Number of tosses = \(\frac{1}{2}\) * 24 = 12

So, Answer will be B
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

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Re: If all the coins are fair, in that the probability of heads = probab [#permalink]
I think this should be rephrased a little bit. It would be better to state that she flipped one coin 24 times. As of now, one may assume that she flipped all 24 coins at the same time in which case the solution would be different.
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Re: If all the coins are fair, in that the probability of heads = probab [#permalink]
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