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Probability Online Practice Questions :Aptitude Questions

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This Aptitude Test Questions sections presents "Probability" solved problems. Probability Questions are provided with detailed answers to every question.

Online practice of these simple Aptitude Problems on PROBABILITY with solution will help those students who are aiming for competitive exams like Bank PO, Bank Clerk, SBI PO, PNB Bank Exam or various Teaching Exams.

Instructions
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  • Click on options to check your answer.
  • Use the timer on the right if want to calculate time taken in solving the questions.

1.

A number X is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What is the probability that $|X| < 2$

A.

5/7

B.

3/7

C.

3/5

D.

1/3

View Ans

 

Answer – (B)

Solution:

X can take 7 values.
To get $|X| < 2$ ( i.e., $-2 < X < +2$) take $X= \{-1, 0, 1\}$

⇒ $P(|X| < 2) = \text" Favourable Cases"/\text"Total Cases"$
= 3/7

2.

Two brother X and Y appeared for an exam. The probability of selection of X is 1/7 and that of B is 2/9. Find the probability that both of them are selected.

A.

1/63

B.

2/35

C.

2/63

D.

9/14

View Ans

 

Answer – (C)

Solution:

Let A be the event that X is selected and B is the event that Y is selected.

$P(A)= 1/7, P(B) = 2/9$.

Let C be the event that both are selected.

$P(C)= P(A)×P(B)$ as A and B are independent events:
$= (1/7) × (2/9)$
= 2/63

3.

An urn contains 6red, 5 blue and 2 green marbles. If 2 marbles are picked at random, what is the probability that both are red?

A.

6/13

B.

5/26

C.

5/13

D.

7/26

View Ans

 

Answer – (B)

Solution:

P(Both are red) $= {^6C_2}/{^13C_2}$
= 5/26

4.

Four dice are thrown simultaneously. Find the probability that all of them show the same face.

A.

1/216

B.

1/36

C.

4/216

D.

3/216

View Ans

 

Answer – (A)

Solution:

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:
$= 6×6×6×6= 6^4$
$n(S)=6^4$

Let X be the event that all dice show the same face.
X $= \{(1,1,1,1,),(2,2,2,2),(3,3,3,3),(4,4,4,4),(5,5,5,5),(6,6,6,6)\}$
$n(X)= 6$

Hence required probability $= {n(X)}/{n(S)} = 6/{6^4}$
= 1/216

5.

A bag contains 12 white and 18 black balls. Two balls are drawn in succession without replacement. What is the probability that first is white and second is black?

A.

18/145

B.

18/29

C.

36/135

D.

36/145

View Ans

 

Answer – (D)

Solution:

The probability that first ball is white:
$= {^12C_1}/{^30C_1}= 12/30 =2/5$

Since, the ball is not replaced; hence the number of balls left in bag is 29.
Hence the probability the second ball is black:
$={^18C_1}/{^29C_1} = 18/29$

Required probability $= (2/5) × (18/29)$
= 36/145