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

• chwala

#### chwala

Gold Member
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.

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}##

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

WWGD and chwala
chwala said:
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
Mark44 said:
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...

chwala said:
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.

chwala said:
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.

chwala
chwala said:
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.

chwala