A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pi

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

(1/2^11)*(999/1000) + 0.5/1000 = 2023/2048000?

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!

Posted from my mobile device

Harsh9676 The ques says the coin is double headed.

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

Could anyone please provide explanation.
I am also getting this answer.
TIA

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

Posted from my mobile device

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.

Posted from my mobile device

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.

Source:-Quora