Answer:
Probability of graduating this semester is 0.7344
Step-by-step explanation:
Given the data in the question;
let A represent passing STAT-314
B represent passing at least in MATH-272 or MATH-444
M1 represent passing in MATH-272
M2 represent passing in MATH-444
C represent passing GERMAN-32
now
P( A ) = 0.85, P( C ) = 90, P( M1 ) = P( M2 ) = 0.8
P( B ) = P( pass at least one of either MATH-272 or MATH-444 ) = P( pass in MATH-272 but not MATH-444 ) + ( pass in MATH-444 but not in MATH 272) + P( pass both )
P( B ) = P( M1 ) × ( 1 - P( M2 ) ) + ( 1 - P( M1 ) ) × P( M2 ) + P( M1 ) × P( M2 )
we substitute
⇒ 0.8×0.2 + 0.2×0.8 + 0.8×0.8 = 0.16 + 0.16 + 0.64 = 0.96
∴ the probability of graduating this semester will be;
⇒ P( A ) × P( B ) × P( C )
we substitute
⇒ 0.85 × 0.96 × 90
⇒ 0.7344
Probability of graduating this semester is 0.7344
