Respuesta :
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

Answer with explanation:
Universal set =U=26
B=8
Y=12
B ∩ Y=8
Probability of an event is defined as total favorable outcome divided by total possible outcome.
[tex]P(B)=\frac{8}{26}\\\\P(Y)=\frac{12}{26}\\\\ P(B\cap Y)=\frac{8}{26}[/tex]
[tex]P(\frac{B}{Y})=\frac{P(B\cap Y)}{P(Y)}\\\\P(\frac{B}{Y})=\frac{\frac{8}{26}}{\frac{12}{26}}\\\\P(\frac{B}{Y})=\frac{8}{12}\\\\P(\frac{B}{Y})=\frac{2}{3}[/tex]
=0.666.....
=0.67 (approx)→→Required probability