Wells Fargo Probability Problems with Solutions Quiz-1

Solution –

Out of 30 items, two can selected in 30C2 ways .

Thus, Total elementary events = 30C2 .

Consider the events :

A = Getting two apples

B = Getting two good items

Required Probability is :

P(A∪B) = P(A) + P(B) – P(A∩B) …(i)

There are 20 apples, out of which 2 can drawn in 20C2 ways .

P(A) = 20C2/30C2

There are 8 defective items and 22 are good, Out of 22 good items, two can be can drawn in 22C2ways .

P(B) = 22C2/30C2

Sice there are 15 items which are good apples, out of which 2 can be selected in 15C2 ways .

P(A∩B) = 15C2/30C2

Substituting the values of P(A), P(B) and P(A∩B) in (i)

Required probability is = (20C2/30C2) + (22C2/30C2) – (15C2/30C2)

= 316/435

Solution:

Consider the two events:

A = Person gets electric contract

B = Person gets plumbing contract

We have,

P(A) = 2/5

P(B’) = 4/7

P(A∪B) = 2/3

Now,

P(A∩B) = P(A) + P(B) – P(A∪B)

= (2/5) + (1 – 4/7) – (2/3)

= 17/105

Solution : S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Favorable outcomes = {HTT, THT, TTH, TTT}

Total number of outcomes = 8

Number of favorable outcomes = 4

Required probability = 4 / 8 = 0.50

Solution : Total number of outcomes = 52

Number of favorable outcomes = Number of red cards = 26

=> Required probability = 26 / 52 = 0.50

Solution : Outcome will be favorable if the two balls drawn are of the same color.

=> Number of favorable outcomes = 6 C 2 + 4 C 2 = 21

Total number of outcomes = 10 C 2 = 45

Therefore, required probability = 21 / 45 = 7 / 15

Solution : S = {1, 2, 3, 4, 5, 6}

Favorable outcomes = {2, 4, 6}

Required probability = 3 / 6 = 0.50

Solution : Total number of outcomes = 20 C 2 = 190

Number of favorable outcomes = 10 C 1 x 10 C 1 = 100

Therefore, required probability = 100 / 190 = 10 / 19

Solution:

Let E1, E2, E3 and A be the defined events as follows :

E1 = The bulb is manufactured by machine A

E2 = The bulb is manufactured by machine B

E3 = The bulb is manufactured by machine C

A = The bulb is defective

According to given conditions ;

P(E1) = 25/100

P(E2) = 35/100

P(E3) = 40/100

Now, probability that the bulb is defective given that is produced by Machine A

P(A/E1) = 5/100

and, probability that the bulb is defective given that is produced by Machine B

P(A/E2) = 4/100

and, probability that the bulb is defective given that is produced by Machine C

P(A/E3) = 2/100

Using total law of probability, we have

P(A) = P(E1)P(A/E1) + P(E2)P(A/E2) + P(E3)P(A/E3)

= (25/100)(5/100) + (35/100)(4/100) + (40/100)(2/100)

= 0.0345

Hence, the probability that the bulb is defective is 0.0345

A red ball can be drawn in two ways:

Selecting bag I and then drawing a red ball from it .

Selecting bag II and then drawing a red ball from it .

Let E1, E2 and A be the defined events as follows :

E1 = Selecting bag I

E2 = Selecting bag II

A = Drawing red ball

Since selecting one of the two bags at random .

P(E1) = 1/2

P(E2) = 1/2

Now, probability of drawing red ball when first bag has been chosen

P(A/E1) = 4/7

and, probability of drawing red ball when second bag has been chosen

P(A/E2) = 2/6

Using total law of probability, we have

P(A) = P(E1)P(A/E1) + P(E2)P(A/E2)

= (1/2)(4/7) + (1/2)(2/6)

= 19/42

Hence, the probability of drawing a red ball is 19/42

Solution:

Consider the two events:

A = Person gets electric contract

B = Person gets plumbing contract

We have,

P(A) = 2/5

P(B’) = 4/7

P(A∪B) = 2/3

Now,

P(A∩B) = P(A) + P(B) – P(A∪B)

= (2/5) + (1 – 4/7) – (2/3)

= 17/105