0
$\begingroup$

Problem is here 2(d)
Original Solution is here 2(d)
My approach:
Let A be the event that at least one roll results in a 3

$$P(A)=1−P(no\ rolls\ resulted\ in\ 3)=1− (3/4)^6$$

Now let K be the random variable representing the number of 3’s in the 6 rolls. The (unconditional)PMF $p_K(k)$for K is given by

$$p_K(k)\ = \ \binom {6}{k} \biggl(\frac{1}{4}\biggl)^k \biggl(\frac{3}{4}\biggl)^{6-k}$$

Till now the solution looks same but now in next steps am little bit confused.
We are given at least one roll resulted in 3 => 5 rolls are left
Probability of at least one roll resulted in 3 is P(A)
Probability of getting k 3's given that at least one roll resulted in 3 should be:
$$p_{K|A}(k)\ =\ P(A) * \binom {5}{k} \biggl(\frac{1}{4}\biggl)^k \biggl(\frac{3}{4}\biggl)^{5-k}\quad \quad ;k=1,2,3,4,5$$ First we ensure that one roll is 3 then we go for next rolls.
Am confused here is it correct if not then why?

And another confusion is:
We are given at least one roll resulted in 3.
And are asked for conditional PMF of the number of 3’s given at least one roll resulted in 3.
Event of getting one 3 given that one roll resulted in 3 has probability of 1. i.e. $p_{K|A}(1)\ = 1$ and $\sum_{k} p_K(k) = 1$
so $p_{K|A}(k)\ = 0 \quad \quad ; k = 2,3,4,5,6$
But it's not the case. So where am I wrong?

$\endgroup$
  • $\begingroup$ "... of THIS die.." What type of die is "this" die? Your attempt seems to imply that it is a fair four-sided die where exactly one of the sides shows the number $3$. This should be made explicitly clear in your post. $\endgroup$ – JMoravitz May 2 at 15:12
  • $\begingroup$ Sorry for the that. I corrected that mistake. $\endgroup$ – yuvraj97 May 2 at 16:09
0
$\begingroup$

There are 2 errors in your thinking, a small one and a bigger, conceptual one.

The small one is that your conditional PMF should look like this

$$p_{K|A}(k)\ =\ P(A) * \binom {5}{k-1} \biggl(\frac{1}{4}\biggl)^{k-1} \biggl(\frac{3}{4}\biggl)^{6-k}\quad \quad ;k=1,2,3,4,5,6$$

It's still incorrect (because of the big error), but if you want exactly $k$ $3$'s and you already "have one $3$", then you need to find the remaining $k-1$ $3$'s. This error is a simple shift "$k \leftrightarrow k-1$".

The big, conceptual error is in your line

We are given at least one roll resulted in 3 => 5 rolls are left

The problem is that with the exception of the $p(A)$ calculation, the remaining formula assumes that you know that a specific roll (say the first) was a $3$. When you say "$5$ rolls are left", the following formula assumes that you can pinpoint the 5 remaining rolls. But you can't do that, because there might be (or are even required to be for $k> 1$) several rolls that gave a $3$, so which one are you taking as the 'set' one'?

An example with less dice might be more intuitive: Take two classical $6$-sided fair dice, a red one and a green one. You roll both (independent results) and want a sum of $12$.

If you can see only the red one immediately after the roll (the green one fell off the table, for example) and the red one shows a $6$, what's your probability

$$P(\text{red+green}=12|\text{red}=6)?$$

The answer should be intuitive and correct: You need a $6$ on the green die and because of independence, that happens with probability $\frac16$:

$$P(\text{red+green}=12|\text{red}=6) = \frac16$$

Now on your next try, both dice fall off the table (you can't see them) and someone with red-green blindness enters the room, sees one die and tells you it's a $6$. Even they don't know if it's the red one or green one. So now the question is, what is

$$P(\text{red+green}=12|\text{red}=6 \text{ or green}=6)?$$

The intuitive answer might be the same as before: You have one $6$, the other die needs to be a $6$ as well, since independence that happens with probability $\frac16$.

But that's wrong!

If we record the result of the dice throws as (r,g), meaning the reusult of the red die first and the green die second, there are $36$ equally probable outcomes: $(1,1), (1,2),\ldots,(1,6),(2,1)\ldots, (6,5), (6,6)$.

In the first scenario (you saw the red die with $6$), the $36$ outcomes where reduced to $6$ (still equally possible) results:

$$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$$

One of those results leads to the sum of $12$, so the probability of that is now $\frac16$.

In the second scenario (the red-green blind person observed one die to be a $6$), the $36$ outcomes were reduced to $11$ (still equally possible) results:

$$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6),(5,6),(4,6),(3,6),(2,6),(1,6)$$

Again, only one outcome gives the result of sum $12$, so the probability under question is

$$P(\text{red+green}=12|\text{red}=6 \text{ or green}=6)=\frac1{11}$$

The fact that you don't know which die is the $6$ changes the picture. If there are multiple ways to choose and fix the roll of the 'known' result, your thinking will lead to incorrect results.

$\endgroup$
  • $\begingroup$ Now I get it I stuck with the turns. Thank you very much for clarification. $\endgroup$ – yuvraj97 May 4 at 0:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.