Twice the difference of a number squared and 7
1 answer:
You might be interested in
1. B and Y are 2 events,and Y has already occurred.
2. P(B|Y)=P(B∩Y)/P(Y),
gives the probability of B to occur, once Y has already happened.
3. According to the givens (check the diagram at the bottom, of the described problem),
P(B∩Y)=n(B∩Y)/n(U)=8/26=4/13
P(Y)=n(Y)/n(U)=12/26=6/13
P(B|Y)=P(B∩Y)/P(Y)=(4/13)/(6/13)=4/6=2/3=0.67
2. P(B|Y)=P(B∩Y)/P(Y),
gives the probability of B to occur, once Y has already happened.
3. According to the givens (check the diagram at the bottom, of the described problem),
P(B∩Y)=n(B∩Y)/n(U)=8/26=4/13
P(Y)=n(Y)/n(U)=12/26=6/13
P(B|Y)=P(B∩Y)/P(Y)=(4/13)/(6/13)=4/6=2/3=0.67