How do I solve this probability question? (Pic. included)

Question

anonymous · Answer

attachment

anonymous · Answer

P(A union B) = P(A) + P(B)
0.8 = 05.+ P(B)
P(B) = 0.3

anonymous · Answer

P(B)+P(C)-P(B)P(C)
= 0.3 +P(C) - (0.3) P(C)

amistre64 · Answer

venn diagram perhaps?

amistre64 · Answer

can we assume c and a are exclusive ?

anonymous · Answer

yes

amistre64 · Answer

P(BuC) = P(B) + P(C) - P(BnC)

amistre64 · Answer

.3 + P(C) - P(BnC) = .75

P(C) - P(BnC) = .45

 P(BnC) = P(C) - .45

amistre64 · Answer

i wonder

 P(B) * P(C) = P(C) - .45

 P(B) * P(C) - P(C)=  - .45

 P(C) ( P(B) - 1)=  - .45

 P(C) =   .45/.7 = .64  which is prolly not the correct way to do it :)

amistre64 · Answer

hmmm, but it is an answer option isnt it ....

or, since we have options can we work it backwards to be sure?

amistre64 · Answer

pfft, i like it; but then again i cant be too sure ;)

kropot72 · Answer

$$P(A \cup B)=P(A)+P(B)$$
$$0.8=0.5+P(B)$$
$$P(B)=0.3$$
$$P(B \cup C)=P(B)+P(C)-P(BintersectionC)$$
$$0.75=0.3+P(C)-P(B)\times P(C)$$
$$0.45=P(C)-0.3P(C)$$
$$P(C)=\frac{0.45}{0.7}=0.643$$