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S1 discrete random variables Q help.

http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/January%202011%20-%20QP/6683_01_que_20110114.pdf

Q6 e,f,g
for a, showed k =0.1
b. E(X)=3
c E(X2)=10
d. Var(2-5X)=25, as Var(X)=1
e. i know that to get a value of 4, you need to have a combination of (1,3) and (2,2), what should i do next? thanks
f. Stuck
g. Stuck

Thanks :smile:

+rep too :smile:
Original post by Hi, How are you ?
http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/January%202011%20-%20QP/6683_01_que_20110114.pdf

Q6 e,f,g
for a, showed k =0.1
b. E(X)=3
c E(X2)=10
d. Var(2-5X)=25, as Var(X)=1
e. i know that to get a value of 4, you need to have a combination of (1,3) and (2,2), what should i do next? thanks
f. Stuck
g. Stuck

Thanks :smile:

+rep too :smile:


For e, you need to consider all the different ways of making 4. You are correct in thinking you need combinations of (1,3) and (2,2). Using these "combinations", how many different ways can you add them to get the number 4?

EDIT: Then for 'f', you use a similar thing. Let me explain how they got one.

Let's look at how they got P(X1+X2=2)=0.01 P(X_1 + X_2 = 2) = 0.01 .

The only way you can make 2 = 1 + 1, as you can't get 2 + 0 (as you can't get 0). In other words, you want 1 AND 1, i.e P(X = 1) * P(X = 1) = 0.1 * 0.1 = 0.01.

Now, you use this similar method to get the rest.

EDIT: For 'g', you need to think things through a little. You want the probability of P(1.5<X1+X23.5 P(1.5 < X_1 + X_2 \leq 3.5 .

Now, you know quite clearly that you can't get probability P(X = something . 5). So you only look at integers. Does this make a little more sense in what to do?
(edited 11 years ago)
Original post by claret_n_blue
For e, you need to consider all the different ways of making 4. You are correct in thinking you need combinations of (1,3) and (2,2). Using these "combinations", how many different ways can you add them to get the number 4?

EDIT: Then for 'f', you use a similar thing. Let me explain how they got one.

Let's look at how they got P(X1+X2=2)=0.01 P(X_1 + X_2 = 2) = 0.01 .

The only way you can make 2 = 1 + 1, as you can't get 2 + 0 (as you can't get 0). In other words, you want 1 AND 1, i.e P(X = 1) * P(X = 1) = 0.1 * 0.1 = 0.01.

Now, you use this similar method to get the rest.

EDIT: For 'g', you need to think things through a little. You want the probability of P(1.5<X1+X23.5 P(1.5 < X_1 + X_2 \leq 3.5 .

Now, you know quite clearly that you can't get probability P(X = something . 5). So you only look at integers. Does this make a little more sense in what to do?


hi, so for e, I done (0.1*0.3)+(0.2*0.2)=0.07, what did i do wrong?
Original post by Hi, How are you ?
hi, so for e, I done (0.1*0.3)+(0.2*0.2)=0.07, what did i do wrong?


There are two ways of getting 4 with 1 and 3.

You either get 1 first and then 3 OR 3 first and then 1.
Original post by claret_n_blue
There are two ways of getting 4 with 1 and 3.

You either get 1 first and then 3 OR 3 first and then 1.


Doesn't that mean they are 2 ways for getting (2,2), if you get what I mean, even though they are the same values?
Original post by Hi, How are you ?
Doesn't that mean they are 2 ways for getting (2,2), if you get what I mean, even though they are the same values?


No. Because they're the same value there's not two ways. If you think of it as children and let's say we are looking the different ways of either getting 1 boy and 1 girl or 2 boys. There are two ways of getting one boy and 1 girl. Either the boy is born first, then the girl, or the girl is born first and then the boy. However, there is only 1 way of getting the two boys, first one boy is born, then the other.

Similarly, there is only way of making 4 from 2 and 2 and that is simply 2 + 2 = 4. Order here doesn't matter as it will still be the same thing

With 1 and 3 however, either you get 1 first then 3 or 3 first then 1.
Original post by claret_n_blue
No. Because they're the same value there's not two ways. If you think of it as children and let's say we are looking the different ways of either getting 1 boy and 1 girl or 2 boys. There are two ways of getting one boy and 1 girl. Either the boy is born first, then the girl, or the girl is born first and then the boy. However, there is only 1 way of getting the two boys, first one boy is born, then the other.

Similarly, there is only way of making 4 from 2 and 2 and that is simply 2 + 2 = 4. Order here doesn't matter as it will still be the same thing

With 1 and 3 however, either you get 1 first then 3 or 3 first then 1.


I see, thanks, can't rep at the moment, sorry :colondollar:

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