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Bunuel
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A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pi 15 Jan 2021, 08:15
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A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pick a coin at random, and toss it 10 times. Given that you see 10 heads, what is the probability that the next toss of that coin is also a head?


A. \(\frac{2,023}{1,024,000}\)

B. \(\frac{1}{1,000} \)

C. \(\frac{253}{500}\)

D. \(\frac{7,531}{10,000}\)

E. \(\frac{1,023}{1,024}\)


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In a box there are 1000 coins, of which 999 are fair and 1 is 2-headed (unfair). Mark picks up one coin from the box randomly and throws it 10 times and gets 10 heads in a row. What is the probability that he has picked the unfair coin?


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A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pi 15 Jan 2021, 22:55
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Bunuel
A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pick a coin at random, and toss it 10 times. Given that you see 10 heads, what is the probability that the next toss of that coin is also a head?


A. \(\frac{2,023}{1,024,000}\)

B. \(\frac{1}{1,000} \)

C. \(\frac{253}{500}\)

D. \(\frac{7,531}{10,000}\)

E. \(\frac{1,023}{1,024}\)
Show more


Hi Bunuel. The question says pick a coin and toss it 10 times. This would indicate that the same coin is tossed 10 times without replacement and therefore it will not affect the probability of the 11th toss as we know that the previous 10 tosses are heads and the probability of those is = 1.

Then the answer should be \(\frac{1}{1000} * 1 + \frac{999}{1000} * \frac{1}{2} = \frac{1001}{2000}\).


Since this is not among the options, then should we assume that they are with replacement and the 11th toss is affected by the previous 10 tosses. In that case shouldn't the word 'it' be replaced?


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A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pi 16 Jan 2021, 02:36
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Bunuel
A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pick a coin at random, and toss it 10 times. Given that you see 10 heads, what is the probability that the next toss of that coin is also a head?


A. \(\frac{2,023}{1,024,000}\)

B. \(\frac{1}{1,000} \)

C. \(\frac{253}{500}\)

D. \(\frac{7,531}{10,000}\)

E. \(\frac{1,023}{1,024}\)


Hi Bunuel. The question says pick a coin and toss it 10 times. This would indicate that the same coin is tossed 10 times without replacement and therefore it will not affect the probability of the 11th toss as we know that the previous 10 tosses are heads and the probability of those is = 1.

Then the answer should be \(\frac{1}{1000} * 1 + \frac{999}{1000} * \frac{1}{2} = \frac{1001}{2000}\).


Since this is not among the options, then should we assume that they are with replacement and the 11th toss is affected by the previous 10 tosses. In that case shouldn't the word 'it' be replaced?


Arun Kumar
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You pick one coin at random at toss it 10 times (the same coin 10 times). The question asks the probability that the next toss of that coin is also a head.
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A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pi 16 Jan 2021, 21:05
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Bunuel

You pick one coin at random at toss it 10 times (the same coin 10 times). The question asks the probability that the next toss of that coin is also a head.
Show more


Hi Bunuel. Then in that case shouldn't the answer be \(\frac{1}{1000} * 1 + \frac{999}{1000} * \frac{1}{2} = \frac{1001}{2000}\)
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A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pi 16 Jan 2021, 21:25
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CrackVerbalGMAT
Bunuel

You pick one coin at random at toss it 10 times (the same coin 10 times). The question asks the probability that the next toss of that coin is also a head.


Hi Bunuel. Then in that case shouldn't the answer be \(\frac{1}{1000} * 1 + \frac{999}{1000} * \frac{1}{2} = \frac{1001}{2000}\)
Show more


Hey Team,

I did not get the answer but going through your explanation, I got a doubt.

How are you concluding that the double sided coin is a coin with two heads? It can be coin with two tails.

Pls help.

Thanks
Harsh
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A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pi 16 Jan 2021, 22:54
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(1/2^11)*(999/1000) + 0.5/1000 = 2023/2048000?
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A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pi 17 Jan 2021, 05:09
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Good morning everyone,

Can someone explain how I did wrong?

Chance to have the only one double head is 1/1000. Chance to have the rest fair coins is 999/1000. For the fair coins to toss eleven times and have heads is (1/2)^11.

So, the chance to have eleven heads combine is 1/1000 + (999/1000 x (1/2)^11) = 3047/2048000

Can someone help?

Thank you!

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A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pi 17 Jan 2021, 05:34
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Harsh9676
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Bunuel

You pick one coin at random at toss it 10 times (the same coin 10 times). The question asks the probability that the next toss of that coin is also a head.


Hi Bunuel. Then in that case shouldn't the answer be \(\frac{1}{1000} * 1 + \frac{999}{1000} * \frac{1}{2} = \frac{1001}{2000}\)


Hey Team,

I did not get the answer but going through your explanation, I got a doubt.

How are you concluding that the double sided coin is a coin with two heads? It can be coin with two tails.

Pls help.

Thanks
Harsh
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Harsh9676 The ques says the coin is double headed.
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A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pi 15 Apr 2021, 01:27
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P(double headed) = 1/1000
P(fair coin). = 999/1000

1. First let us find the probability that the flipped coin is double headed.
= P(Double Headed )x1 / {( 1/1000 x 1^10+(999/1000)x (1/2)^10)}= 1024/2023. ~ 0.5


2. P(double headed coin) given 10 tosses were heads = 1024/2023. ~ 0.5
P( fair coin) given 10 tosses were heads = 999/2023. ~ 0.5

3. P( next toss getting Head) = 1x(1024/2023) +(1/2)x(999/2023)= 3047/4046= 0.750389. choice D.
Or = 0.5x1+0.5x 0.5 = 0.75
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A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pi 14 Dec 2021, 10:16
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CrackverbalGMAT
Bunuel

You pick one coin at random at toss it 10 times (the same coin 10 times). The question asks the probability that the next toss of that coin is also a head.


Hi Bunuel. Then in that case shouldn't the answer be \(\frac{1}{1000} * 1 + \frac{999}{1000} * \frac{1}{2} = \frac{1001}{2000}\)
Show more

Could anyone please provide explanation.
I am also getting this answer.
TIA
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A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pi 14 Dec 2021, 12:19
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This question is really asking for the probability that the coin being tossed is fair or unfair.

Let's calculate probability of fairness given the stated pattern:

Probability of 10 heads in a row is (999/1000)*.5^10
+ 1*(1/1000) =

999/(1024*1000) + 1/1000=

(999+1024)/(1024*1000)=

2023/(1024*1000)


This is the denominator.

The numerator is the contribution from the fair coin.

The numerator divided by denominator is the probability of the coin being fair:

(999/(1024*1000))/(2023/(1024*1000)) = 999/2023

The probability of the coin being unfair is 1- the above or

(2023-999)/2023 = 1024/2023

The probability of a head on the next flip is then

(999/2023)*.5 + (1024/2023)*1

= (999/4046) + (1024/2023)
= (999+2048)/4046
= 3047/4046

This is around 75%. Answer D is around 75%.

Is (10000/4046)*3047= 7531 ?

A little calculator work shows that this is exactly true, but the estimate of 75% is all that is needed to establish D as the answer

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A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pi 14 Dec 2021, 13:35
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CrackverbalGMAT
Bunuel

You pick one coin at random at toss it 10 times (the same coin 10 times). The question asks the probability that the next toss of that coin is also a head.


Hi Bunuel. Then in that case shouldn't the answer be \(\frac{1}{1000} * 1 + \frac{999}{1000} * \frac{1}{2} = \frac{1001}{2000}\)
Show more

Another way to think about this is that there are two "events" with uncertainty associated with them, the first being the selection of the coin and second being the series of flips.

These two are combined in the resulting pattern of 10 flips.

The pattern is a mutual weighting of each.

An increasing series of heads weights the probability of the unfair coin in an increasing way and diminishes the weighting of the fair coin.

The proper approach is to disentangle these with Bayes Theorem.

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A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pi 14 Dec 2021, 23:31
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You use Bayes Theorem to find the answer. Let's split problem into two parts:
1) What is the probability you picked the double-headed coin (now referred as D)?
2) What is the probability of getting a head on the next toss?

Question 2 follows very naturally after question 1, so let's tackle question 1.

We are trying to find the probability of having a double-headed coin. We know that the same coin has been flipped 10 times, and we've gotten 10 heads (intuitively, you're probably thinking that there is a significant chance we have the double-headed coin). Formally, we're trying to find P(D | 10 heads).

Using Bayes rule:
P(10 H | D) * P(D)
P(D | 10 H) = ---------------------------
P(10H)

Tackling the numerator, the prior probability, P(D) = 1/1000. If we used the double headed coin, the chance of getting 10 heads, P(10 H | D) = 1 (we always flip heads). So the numerator = 1 / 1000 * 1 = 1 / 1000.

The denominator, P(10H) is just P(10 H | D) * P(D) + P(10 H | Fair) * P(Fair). This makes sense because we are simply enumerating over the two possible coins. The first part of P(10H) is the exact same as the numerator (1 / 1000). Then the second part: P(Fair) = 999/1000. P(10 H | Fair) = (1/2) ^ 10 = 1/1024. Thus P(10 H | Fair) * P(Fair) = .0009756. The denominator then equals .001 + .0009756.

Since we have all the components of P(D | 10 H), compute and you'll find the the probability of having a double headed coin is .506. We have finished the first question.

The second question is then easily answered: we just compute the two individual possibilities and add.

P(H) = P(D) * P(H | D) + P(Fair) * P(H | Fair) = .506 * 1 + (1 - .506) * (.5) = .753.

So there is a 75.3% chance you will flip a heads.

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