1. ## More probability help!!

Just got a couple more questions I need help with on probability, Thank you so much guys.

Let X be a normally distributed variable with mean = u and standard deviation = o. If u < a< b and Pr(x > b) = q
a) Pr(a < x < b)
b) Pr(-b < x < -a)
c) Pr( x > b | x > a)

2. In order to asses the quality of video games being produced the manufacturer selects 10 games at random before they are boxed for delivery, they are inspected.
a) If it is known 8 % of games are defective find:
i) probability that one of the games on the sample will be defective.
ii) that at least one of the games in the sample are defective.
iii) one of the games in the sample will be defective, given that least one of the games are defective

2. Originally Posted by RoboStar
Just got a couple more questions I need help with on probability, Thank you so much guys.

Let X be a normally distributed variable with mean = u and standard deviation = o. If u < a< b and Pr(x > b) = q
a) Pr(a < x < b) Mr F says: = Pr(x < b) - Pr(x < a) = 1 - Pr(x > b) - Pr(x < a) = 1 - q - Pr(x < a).

b) Pr(-b < x < -a) Mr F says: = Pr(x < -a) - Pr(x < -b) = Pr(x > a) - Pr(x > b) = Pr(x > a) - q.

c) Pr( x > b | x > a) Mr F says: $\displaystyle {\color{red}= \frac{\Pr(x > b \, \text{and} \, x > a)}{\Pr(x > a)} = \frac{\Pr(x > b)}{\Pr(x > a)}}$.

Mr F says: Unless more is known about a eg. Pr(x > a) = .... or Pr(x < a) = .... there's not much more can be done here .....

2. In order to asses the quality of video games being produced the manufacturer selects 10 games at random before they are boxed for delivery, they are inspected.
a) If it is known 8 % of games are defective find:
i) probability that one of the games on the sample will be defective.
ii) that at least one of the games in the sample are defective.
iii) one of the games in the sample will be defective, given that least one of the games are defective

2. Let X be the random variable number of defective video games in sample.

I suppose you're to assume sampling from a large population ....? Then X ~ Binomial(n = 10, p = 0.08).

i) Pr(X = 1) = ....

ii) $\displaystyle \Pr(X \geq 1) = 1 - \Pr(X = 0) = ....$

iii) $\displaystyle \Pr(X = 1 \, | \, X \geq 1) = \frac{\Pr(X = 1 \, \text{and} \, X \geq 1)}{\Pr(X \geq 1)} = \frac{\Pr(X = 1)}{\Pr(X \geq 1)} = ....$