Question

Two students have a bitter disagreement about the answer to a homework question and decide to settle their differences in a duel at the golf cours. The rules of the duel are simple:
Each student takes a single shot from the 18th tee. If only one student's shot lands on the green, that student is declared the victor and the duel is over. If both students' shots land on the green, the result is a draw and the duel is over. If they both miss the green, then they repeat another round (and so on) until the duel ends in a draw or with a victor. The result of each tee shot is independent. Student A has not spent much time at the golf course, resulting in a 0.55 probability that her tee shot will hit the green. Student B golfs regularly, resulting in a 0.76 probability that his tee shot will hit the green.

Required:
a. Whatâs the probability that the duel ends in round 1?
b.Whatâs the probability that the duel ends in the 3rd round?
c. Whatâs the probability that the duel ends in a draw (i.e., that neither student wins)?

Answer 1

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

Step-by-step 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