1. ## Unfail Coin

Hi guys I'm new to this forum, hope to get along with you all.
I'm being having trouble in solving this 3 question

I know how to solve a fair coin just having trouble in solving this one hope you guys can help
Also I'm not sure if I should count Zero in there or not because there is still chance of getting zero Heads

2. Originally Posted by BkNekno
Hi guys I'm new to this forum, hope to get along with you all.
I'm being having trouble in solving this 3 question

I know how to solve a fair coin just having trouble in solving this one hope you guys can help
Let X be the random variable number of heads.

X ~ Binomial(n = 8, p = 0.7)

9. Calculate Pr(X = 2) + Pr(X = 4) + Pr(X = 6) + Pr(X = 8).

9., 10. and 11. Do the calculations using the fact that X ~ Binomial(n = 8, p = 0.7).

Obviously X = 0 must be considered in 10.

3. Originally Posted by mr fantastic

Obviously X = 0 must be considered in 10.
How about for 9 because i recently did an exercise that is a little similar
and the teacher told us that Zero is counted as an event number too

4. Originally Posted by BkNekno
How about for 9 because i recently did an exercise that is a little similar
and the teacher told us that Zero is counted as an event number too
There is some debate about whether zero is even or odd. If your teacher says to treat it as even, then do so.

5. that is what i though, Zero supposed to be nor possitive nor negative so it can't be odd nor even. Maybe i should show that problem to my school math department dean and have him correct me of the teacher.

Thank you for you fast post

6. Originally Posted by BkNekno
that is what i though, Zero supposed to be nor possitive nor negative so it can't be odd nor even. Maybe i should show that problem to my school math department dean and have him correct me of the teacher.

Thank you for you fast post
Note: I typed a bit too quickly. Please re-read my previous post. There's some debate and it's not an open and shut case - you should google

zero even

and consider the different arguments for an against that have been given.

This thread may well become a hot bed of opinion .....

7. so if zero is counted
if i toss 8 times a fair coin then P(X is even )= {0,2,4,6,8,} = 5/9 ?

8. Originally Posted by BkNekno
so if zero is counted
if i toss 8 times a fair coin then P(X is even )= {0,2,4,6,8,} = 5/9 ?
No. The outcomes are NOT equally likely.

You have to calculate the probabilities using the following:
Originally Posted by mr fantastic
Let X be the random variable number of heads.

X ~ Binomial(n = 8, p = 0.7)
[snip
So $\displaystyle \Pr(X = x) = {8 \choose x} (0.7)^x (0.3)^{8-x}$. Do you realise this ....? So:

$\displaystyle \Pr(X = 0) = (0.3)^8$.

Pr(X = 2) = ....
Pr(X = 4) = ....
Pr(X = 6) = ....

(these are left for you to calculate)

$\displaystyle \Pr(X = 8) = (0.7)^8$.