Find the following probabilities: (A) Events A and B are mutually exclusive events defined on a common sample space. If P (A) = 0.4 and P(A or B) = 0.65, find P(B).

P(A or B) = P(A) + P(B) 0.65 = 0.4 + P(B)

=> P(B) = 0.65 – 0.4 => P(B) = 0.25

(B) Events A and B are defined on a common sample space. If P(A) = 0.30, P(B) = 0.50, and P(A or B) = 0.60, find P(A and B)

P(A or B) = P(A) + P(B) - P(A and B) 0.60 = 0.30 + 0.50 - P(A and B) P(A and B) = 0.80 – 0.60 = 0.20

2. A bag of jelly belly candies contains the following colored jelly beans: red (9), blue (2), orange (3), brown (18), green (8), and yellow (11). Construct the probability distribution for x.

Total number of jelly beans = 9 + 2 + 3 + 18 + 8 + 11 = 51

Probability distribution for x. x P(x) Red 9/51 = 3/17 Blue 8/51 Orange 3/51 = 1/17 Brown 18/51 = 6/17 Green 8/51 Yellow 11/51

3. Find the mean and standard deviation of the following probability distribution: x 1 2 3 P(x) 0.3 0.5 0.2

Mean = 1*0.3 + 2*0.5 + 3*0.2 = 0.3 + 1 + 0.6 = 1.9 Variance= 0.3*(1-1.9)^2 + 0.5*(2-1.9)^2 + 0.2*(3-1.9)^2 = 0.49 Standard deviation = √0.49 = 0.7

5. In testing a new drug, researchers found that 1% of all patients using it will have a mild side effect. A random sample of 12 patients using the drug is selected. Find the probability that: (A) exactly two will have this mild side effect (B) at least one will have this mild side effect.

Here n = 12, p = 0.01 (A) exactly two will have this mild side effect

P(x = 2) = C(12, 2)*(0.01^2)*(0.99^10) = 66*(0.01^2)*(0.99^10) ≈ 0.00597

(B) at least one will have this mild side effect

P(x ≥ 1) = 1 – P(x = 0)

= 1 - C(12, 0)*(0.01^0)*(0.99^12) = 1 - 1*(0.01^0)*(0.99^12) = 1 - 0.8864 ≈ 0.1136