Question

the probability that a student passes subject a, b or c is 98%. the probability that he or she passes a is 41%, b is 59%. the probability that he or she passes a and c is 25% and b and c is 20%. the probability that he or she passes all the 3 subjects is 14%. what is the probability that he or she passes subject c?

Answer 1

Using De Morgan's Laws, we have

P(a+b+c) = P(a)+P(b)+P(c)-P(a*b)-P(b*c)-P(a*c)+P(a*b*c)

P(a+b+c) = 98% = 0.98
P(a) = 41% = 0.41
P(b) = 59% = 0.59
P(c) = need to find
P(a*b) is assumed to be 0 as it is not given.
P(a*c) = 25% = 0.25
P(b*c) = 20% = 0.20
P(a*b*c) = 14% = 0.14

Substituting these values, we get

0.98 = 0.41+0.59+ P(c) - 0.25 - 0.20 + 0 + 0.14 = 1.14 - 0.45 + P(c)

0.98 = 0.69 + P(c)

or P(c) = 0.98-0.69 = 0.29.

Answer 2

Answer:

Step-by-step explanation:

Using De Morgan's Laws,

P(a+b+c) = P(a)+P(b)+P(c)-P(a x b)-P(b x c)-P(a x c)+P(a x b x c)

P(a+b+c) = 98% or 0.98

P(a) = 41% or 0.41

P(b) = 59% or 0.59

P(c) = ?

P(a x b) is assumed to be= 0 

P(a x c) = 25% = 0.25

P(b x c) = 20% = 0.20

P(a x b x c) = 14% = 0.14

Putting the values we get

0.98 = 0.41+0.59+ P(c) - 0.25 - 0.20 + 0 + 0.14

0.98 = 1.14 - 0.45 + P(c)

0.98 = 0.69 + P(c)

P(c) = 0.98-0.69

= 0.29.