S1 maths NEED HELP!!!!!!!!!!!!

for unbiased p(head)=3/4

I'm not asking for p(head) - I'm asking for p(get 1 head and 1 tails out of 2 throws).

Also p(head) doesn't mean anything when you are throwing an unbiased coin twice - it is ambiguous.

can you explain from tree diagram

Fantastic, my next step was to draw a tree diagram for you but you did it yourself which is good.

Now, there are 3 events. One is that you get two heads. Another is that you get two tails. The last one is that you get one heads and one tails - what is the probability of this?

There is two cases for getting one head and one tail
Case 1
Head then tail
Case 2
Tail then head

Probability of case 1
0.5 x 0.5
For case 2
0.5 x 0.5

If we add them0.5×0.5+0.5×0.5 =0.5

Brilliant, I think the point of this example might be lost in the pages but you have worked out that there are two cases (this is why I asked you about the biased example first, because it would be more clear than adding 0.5*0.5 to itself)
So can you see that there are two cases in the original question, one where X1 = 2 and X2 = 3 and the other where X1 = 3 and X2 = 2?

Yeah. .............

i didn't understand this (but it's impossible to pick 2, hence it is impossible to get a total of 5 in 2 goes ) could you explain a bit more please

Yeah I understood

Thanks

No problem.

Are you fine with the rest of the questions?

No problem.

Are you fine with the rest of the questions?

How to do rest of questions

You tell me...

i know how to do part f but no idea about e

What are the range of values that X1 + X2 can take? (Answer: 0,1,2,3,4,6, not 5)

You found the probability for 5 in part d.. can you do it for the rest of the numbers?

for 0
p(x1=0) x p(x2=0) +p(x2=0)x p(x1=0)=0 =0.5

as p(x=0) =2k
2(0.25)=0.50

is that right

Yes

but mark scheme shows 0.25