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For B, note that P(x < 2) = P(x = 0) + P(x = 1) since these are the only two values the random variable can take below 2.
So do i have to do the same thing i did for part a but repeat it for x = 0 and x = 1?
Yep, since you know how to calculate the probability the variable is equal to something you can use that to find the probability it is less than something.
okay and for part c i do tht for x = 3 all the way up to 9? that would take forever...
For part C you can use the fact that P(x > 2) + P(x <= 2) = 1 and use your previous answers to calculate P(x > 2).
Otherwise you're right, it is a bunch of boring calculation :P
I dont understand
If something is not larger than 2, then that means it's less than or equal to 2. So P(x > 2) = 1 - P(x <= 2). From your first two answers you can calculate P(x <= 2) easily so that's why we do it this way.
Im sorry I still dont get it
That's ok, I'm just trying to think of ways to explain it :P Do you know that if you know the probability of something happening then the probability of it not happening is 1 minus that probability?
btw i think i solved part b is it 0.196?
It is :) So do you agree that P(x > 2) = 1 - P(x <= 2)?
I dont understand that formula
It comes from what I said above combined with the fact that is x is not > 2, then it is <= 2.
okay that makes sense
Alright. So can you find P(x <= 2)?
i did c9,1(0.3)^2(0.7)^9-1 and the same this for 0 to get 0.196
i used an online calculator to solve it because i dont know how to do it on my calclator
0.196 is P(x < 2), but you need P(x <= 2)
no part b says just less than 2 and i found out how to do it on my calculator had to use the nCr key
I mean for part C, since we know that P(x > 2) = 1 - P(x <= 2)
so would P(x <=2) be 0.267 + 0.196?
Yes. Then you can find the answer to C
so c would be 1 - (0.267+0.196)?
Wow! thank you I actually understand what you were trying to say earlier
okay so how would i go about doing part d?
is it asking to find x = 3?
Well if 2 <= x <= 4 then that means x can be 2, 3, or 4.
right so how can i approach this?
would i have to solve for x = 2,3,4?
Yeah, I think the easiest way to do it here is to just calculate the probabilities for each case and sum them.
:( okay well atleast I know how to do these srot of questions thank you so much JE!
No worries, glad I could help! :)