Date | November 2013 | Marks available | 4 | Reference code | 13N.2.sl.TZ0.4 |
Level | SL only | Paper | 2 | Time zone | TZ0 |
Command term | Find | Question number | 4 | Adapted from | N/A |
Question
Two events AA and BB are such that P(A)=0.2P(A)=0.2 and P(A∪B)=0.5P(A∪B)=0.5.
Given that AA and BB are mutually exclusive, find P(B)P(B).
Given that AA and BB are independent, find P(B)P(B).
Markscheme
correct approach (A1)
eg 0.5=0.2+P(B), P(A∩B)=00.5=0.2+P(B), P(A∩B)=0
P(B)=0.3P(B)=0.3 A1 N2
[2 marks]
Correct expression for P(A∩B)P(A∩B) (seen anywhere) A1
eg P(A∩B)=0.2P(B), 0.2xP(A∩B)=0.2P(B), 0.2x
attempt to substitute into correct formula for P(A∪B)P(A∪B) (M1)
eg P(A∪B)=0.2+P(B)−P(A∩B), P(A∪B)=0.2+x−0.2xP(A∪B)=0.2+P(B)−P(A∩B), P(A∪B)=0.2+x−0.2x
correct working (A1)
eg 0.5=0.2+P(B)−0.2P(B), 0.8x=0.30.5=0.2+P(B)−0.2P(B), 0.8x=0.3
P(B)=38 (=0.375, exact)P(B)=38 (=0.375, exact) A1 N3
[4 marks]