Answer:
a.) P(duel end in round 1 ) = 0.892
b.) P(duel ends in the 3rd round) = 0.010
c.) P(duel ends in a draw) = 0.418
Explanation:
Let
P(A) - Probability that Student A will hit the green.
P(¬A) - Probability that Student A will miss the green.
P(B) - Probability that Student B will hit the green.
P(¬B) - Probability that Student B will miss the green.
Given,
P(A) = 0.55 , P(B) = 0.76
As we know that,
P(A) + P(¬A) = 1
⇒P(¬A) = 1 - P(A)
= 1 - 0.55
= 0.45
⇒P(¬A) = 0.45
And
P(B) + P(¬B) = 1
⇒P(¬B) = 1 - P(B)
= 1 - 0.76
= 0.24
⇒P(¬B) = 0.24
So, we get
P(A) = 0.55 , P(B) = 0.76
P(¬A) = 0.45, P(¬B) = 0.24
a.)
P(duel end in round 1 ) = P( both hit on green or only one hit on green )
= P(A).P(B) + [ P(A).P(¬B) + P(¬A).P(B) ]
= 0.55(0.76 ) + [ 0.55(0.24) + 0.45(0.76) ]
= 0.418 + [ 0.132 + 0.342 ]
= 0.418 + [ 0.474 ]
= 0.892
⇒P(duel end in round 1 ) = 0.892
b.)
P(duel ends in the 3rd round) = P( duel miss green in 1st round × duel miss green in 2nd round × both hit on green or only one hit on green in 3rd round )
= [ P(¬A).P(¬B) ] × [ P(¬A).P(¬B) ] × { P(A).P(B) + [ P(A).P(¬B) + P(¬A).P(B) ] }
= [ 0.45(0.24) ] × [ 0.45(0.24) ] × { 0.55(0.76 ) + [ 0.55(0.24) + 0.45(0.76) ] }
= [ 0.108 ] × [0.108] × { 0.418 + [ 0.132 + 0.342 ] }
= 0.011664 × 0.892
= 0.010404 ≈ 0.010
⇒P(duel ends in the 3rd round) = 0.010
c.)
P(duel ends in a draw) = P(both students' shots land on the green)
= P(A).P(B)
= 0.55(0.76)
= 0.418
⇒P(duel ends in a draw) = 0.418