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I understand how to get the denominator just fine, but I am missing something on the numerator. I read the answer, but something just isn't clicking.

Thanks!

Welcome to Gmat Club forum.

It would be easier to calculate the probability of opposite event and subtract it from 1. Opposite event: 0 tail, 1 tail, 6 tails.

Probability of getting no tails: \(\frac{1}{2^6}=\frac{1}{64}\);

Probability of getting 1 tail: \(6C1*\frac{1}{2^6}=\frac{6}{64}\), we must multiply by 6C1 or by 6 as tail can occur for any flip from 6, hence in 6 ways;

Probability of getting 6 tails: \(\frac{1}{2^6}=\frac{1}{64}\)

I understand how to get the denominator just fine, but I am missing something on the numerator. I read the answer, but something just isn't clicking.

Thanks!

Welcome to Gmat Club forum.

It would be easier to calculate the probability of opposite event and subtract it from 1. Opposite event: 0 tail, 1 tail, 6 tails.

Probability of getting no tails: \(\frac{1}{2^6}=\frac{1}{64}\);

Probability of getting 1 tail: \(6C1*\frac{1}{2^6}=\frac{6}{64}\), we must multiply by 6C1 or by 6 as tail can occur for any flip from 6, hence in 6 ways;

Probability of getting 6 tails: \(\frac{1}{2^6}=\frac{1}{64}\)

I understand how to get the denominator just fine, but I am missing something on the numerator. I read the answer, but something just isn't clicking.

Thanks!

Welcome to Gmat Club forum.

It would be easier to calculate the probability of opposite event and subtract it from 1. Opposite event: 0 tail, 1 tail, 6 tails.

Probability of getting no tails: \(\frac{1}{2^6}=\frac{1}{64}\);

Probability of getting 1 tail: \(6C1*\frac{1}{2^6}=\frac{6}{64}\), we must multiply by 6C1 or by 6 as tail can occur for any flip from 6, hence in 6 ways;

Probability of getting 6 tails: \(\frac{1}{2^6}=\frac{1}{64}\)

I understand how to get the denominator just fine, but I am missing something on the numerator. I read the answer, but something just isn't clicking.

Thanks!

Welcome to Gmat Club forum.

It would be easier to calculate the probability of opposite event and subtract it from 1. Opposite event: 0 tail, 1 tail, 6 tails.

Probability of getting no tails: \(\frac{1}{2^6}=\frac{1}{64}\);

Probability of getting 1 tail: \(6C1*\frac{1}{2^6}=\frac{6}{64}\), we must multiply by 6C1 or by 6 as tail can occur for any flip from 6, hence in 6 ways;

Probability of getting 6 tails: \(\frac{1}{2^6}=\frac{1}{64}\)

For more on probability and combinatorics please refer to the link: GMAT MATH BOOK

Hi Bunuel,

I understand the numerator part. 2C6 + 3C6 + 4C6 + 5C6 = 56

but how to calculate denominator part. I mean how can i count total no of combinations. I am not getting 64 . Like in normal cases if we calculate for 6 ball, we take 6! as total no of combinations. Please help _________________

I understand how to get the denominator just fine, but I am missing something on the numerator. I read the answer, but something just isn't clicking.

Thanks!

Welcome to Gmat Club forum.

It would be easier to calculate the probability of opposite event and subtract it from 1. Opposite event: 0 tail, 1 tail, 6 tails.

Probability of getting no tails: \(\frac{1}{2^6}=\frac{1}{64}\);

Probability of getting 1 tail: \(6C1*\frac{1}{2^6}=\frac{6}{64}\), we must multiply by 6C1 or by 6 as tail can occur for any flip from 6, hence in 6 ways;

Probability of getting 6 tails: \(\frac{1}{2^6}=\frac{1}{64}\)

For more on probability and combinatorics please refer to the link: GMAT MATH BOOK

Hi Bunuel,

I understand the numerator part. 2C6 + 3C6 + 4C6 + 5C6 = 56

but how to calculate denominator part. I mean how can i count total no of combinations. I am not getting 64 . Like in normal cases if we calculate for 6 ball, we take 6! as total no of combinations. Please help

Each coin can land on heads or tails, so 2 ways. We have 6 coins, so total number of outcomes is 2*2*2*2*2*2 = 2^6. _________________

Re: A fair 2 sided coin is flipped 6 times. What is the [#permalink]

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11 Oct 2015, 21:32

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