Show that ##P(X=2)=\dfrac{2}{15}##

[chwala]

Homework Statement

Pack A consists of $10$ cards numbered $0,0,1,1,1,1,1,3,3,3$. Pack B consists of $6$ cards numbered $0,0,2,2,2,2$. One card is chosen at random from each pack. The random variable $X$is defined as the sum on the two cards.

Show that $P(X=2)=\dfrac{2}{15}$

Relevant Equations

Probability

My approach;

	0	0	1	1	1	1	1	3	3	3
0	0	0	1	1	1	1	1	3	3	3
0	0	0	1	1	1	1	1	3	3	3
2	2	2	3	3	3	3	3	5	5	5
2	2	2	3	3	3	3	3	5	5	5
2	2	2	3	3	3	3	3	5	5	5
2	2	2	3	3	3	3	3	5	5	5

$P(X=2)=\dfrac{8}{60}=\dfrac{4}{30}=\dfrac{2}{15}$

Mark scheme solution;

just sharing...cheers! Any insight is welcome.

 

[chwala]

Part iii. of the question.

Given that $x=3$, find the probability that the card chosen from Pack A is a $1$.

In the markscheme (well understood), they used Conditional probability i.e

Alternatively, from my table above;

We have $P(x=3)=\dfrac {26}{60}$ and the Probability $P(A1∩sum 3)= \dfrac {20}{60}$

therefore;

The probability that the card chosen from Pack A is a $1=\dfrac {20}{60}×\dfrac {60}{26}=\dfrac {20}{26}=\dfrac {10}{13}$

 

[Office_Shredder]

Your answer seems fine to me, the mark scheme says getting the probabilities by counting is fine.

 

[Mark44]

Homework Statement:: Pack A consists of $10$ cards numbered $0,0,1,1,1,1,1,3,3,3$. Pack B consists of $6$ cards numbered $0,0,2,2,2,2$. One card is chosen at random from each pack. The random variable $X$is defined as the sum on the two cards.

Show that $P(X=2)=\dfrac{2}{15}$

A method that requires a lot less writing is this:
To get a total of 2 from the two cards, the only possibility is to select a 0 card from pack A and a 2 card from pack B.
With the r.v. X being the sum of the two cards, P(X = 2) = P(0 from A and 2 from B) = P(0 from A) x P(2 from B) = $\frac 2 {10} \cdot \frac 4 6 = \frac 2 {15}$.

 

[chwala]

A method that requires a lot less writing is this:
To get a total of 2 from the two cards, the only possibility is to select a 0 card from pack A and a 2 card from pack B.
With the r.v. X being the sum of the two cards, P(X = 2) = P(0 from A and 2 from B) = P(0 from A) x P(2 from B) = $\frac 2 {10} \cdot \frac 4 6 = \frac 2 {15}$.

Brilliant! @Mark44 ...

Would one get full marks using this exceptional approach...

 

[Mark44]

Brilliant! @Mark44 ...

Would one get full marks using this exceptional approach...

I don't see why not.
What I did simply uses basic properties of probabilities.

 

[pbuk]

Any insight is welcome.

You haven't got time (or space) to write all that out in an exam. It can easily be seen that the only way to get a sum of 2 is to draw a 0 from the first pack and a 2 from the second pack. There are 2 x 4 = 8 ways to do that and there are 10 x 6 = 60 total draws.

 

[pbuk]

Would one get full marks using this exceptional approach...

On this question? Yes. On the whole exam? Probably not - you won't have time to finish it if you use approaches like this.