Answer to Question #199342 in Statistics and Probability for Bahawal tahir

Question

An oil company is bidding for the rights to drill a well in field A and a well in field B. The probability it will drill a well in field A is 40%. If it does, the probability the well will be successful is 45%. The probability it will drill a well in field B is 30%. If it does, the probability the well will be successful is 55%. Calculate each of the following probabilities: 

a) probability of a successful well in field A, 

b) probability of a successful well in field B, 

c) probability of both a successful well in field A and a successful well in field B, 

d) probability of at least one successful well in the two fields together,

e) probability of no successful well in field A, 

f) probability of no successful well in field B, 

g) probability of no successful well in the two fields together (calculate by two methods), 

h) probability of exactly one successful well in the two fields together. Show a check involving the probability calculated in part h

Answer 1

solution:-

An oil company is bidding for the rights to drill a well in field A and a well in field B. 

Pr( a well in A) =0.40

Pr( Success/well in A)=0.45

Pr( a well in B) =0.30

Pr( Success/well in B)=0.55

a) probability of a successful well in field A

Pr( a successful well in field A) = Pr( a well in A) ×Pr( Success/well in A)

= 0.40×0.45

= 0.18

b) probability of a successful well in field B

Pr( a successful well in field B) = Pr( a well in B) ×Pr( Success/well in B)

= 0.30×0.55

=0.165

c)probability of both a successful well in field A and a successful well in field B

=Pr( both a successful well in field A and a successful well in field B)

=Pr( a successful well in field A) ×Pr( a successful well in field B)

= 0.18×0.165

= 0.0297

d) probability of at least one successful well in the two fields together

=Pr( at least one successful well in two fields)

=Pr[( successful well in field A)U(successful well in field B)]

= 0.18+0.165-0.0297

=0.3152 

e) probability of no successful well in field A

Pr( no successful well in field a)

= Pr( no well in field A) + ( unsuccessful well in field A)

=0.60+(0.40×0.55)

=0.60+0.22

=0.82

f) probability of no successful well in field B

Pr( no successful well in field b)

= Pr( no well in field B) + ( unsuccessful well in field B)

=0.70+(0.30×0.45)

=0.70+0.135

=0.835

g) probability of no successful well in the two fields together 

method1

=Pr( no successful well in field A and no successful well in field B)

=Pr( no successful well in field A) ×Pr( no successful well in field B)

= 0.82×0.835

= 0.6847

method2

=1-{Pr( a successful well in field A and no successful well in field B)+Pr( no successful well in field A and a successful well in field B)+Pr( a successful well in field A and a successful well in field B)}

=1-{Pr( a successful well in field A) ×Pr( no successful well in field B)+Pr( no successful well in field A) ×Pr( a successful well in field B)+Pr( a successful well in field A) ×Pr( a successful well in field B)}

=1-[(0.18×0.835)+(0.82×0.165)+(0.18×0.165)]

=1-[(0.1503)+(0.1353)+(0.0297)]

= 0.6847

h) probability of exactly one successful well in the two fields together.

=Pr( a successful well in field A and no successful well in field B)+Pr( no successful well in field A and a successful well in field B)

=Pr( a successful well in field A) ×

Pr( no successful well in field B)+Pr( no successful well in field A) ×Pr( a successful well in field B)

=(0.18×0.835)+(0.82×0.165)

=(0.1503)+(0.1353)

= 0.2856

check

=1-{Pr( no successful well in field A and no successful well in field B)+Pr( a successful well in field A and a successful well in field B)}

=1-{Pr( no successful well in field A) ×Pr( no successful well in field B)+Pr( a successful well in field A) ×Pr( a successful well in field B)}

=1-[(0.82×0.835)+(0.18×0.165)]

=1-[(0.6847)+(0.0297)]

= 0.2856