Why Does Calculating P(|Y-5| >= 3) Involve Y=7 in a Binomial Distribution?

[Bkid701]

IF Y~B(11, 0.3), find (|Y-5| >= 3)

I got the answer(0.3170) but i don't understand the logic behind this part where i am confused.

can someone explain the working(second working) where i somehow got it blindly correct?


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my working at first:

|Y-5| >= 3
Y >= 8
Y <=2

so P( 2 >= Y >= 8)

P(Y=8) = 0.9994
1 - 0.9994 = 0.0006

P(Y=2) = 0.3127

so P( 2 >= Y >= 8) = 0.3133

However this answer is wrong
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My second working:

my working at first:

|Y-5| >= 3
Y >= 8
Y <=2

so P( 2 >= Y >= 8)

P(Y=7) = 0.9957
1 - 0.9957 = 0.0043

P(Y=2) = 0.3127

so P( 2 >= Y >= 8) = 0.3170

This is the correct working but i don't understand why Y is = 7...
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[haruspex]

|Y-5| >= 3
Y >= 8
Y <=2

so P( 2 >= Y >= 8)

Much clearer to write P( 2 >= Y | Y >= 8)

P(Y=8) = 0.9994

That's the probability that Y <= 8. If you subtract that from 1 you'll have the probability that Y > 8.

 

[Bkid701]

Thank you for your help. I'm getting the idea of it now.

Cheers