3
Q:

A sequence a, ax, ax2, ......, axn, has odd number of terms. Then the median is

 

A) axn2+1 B) axn2-1
C) axn-1 D) axn2

Answer:   D) axn2



Explanation:
Subject: Probability
Exam Prep: Bank Exams
Q:

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

A) 1/216 B) 1/36
C) 2/216 D) 4/216
 
Answer & Explanation Answer: A) 1/216

Explanation:

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:

 

6*6*6*6=64

 

n(S) = 64

 

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)=664 =1216

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48 34473
Q:

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) 1/14
C) 2/63 D) 1/9
 
Answer & Explanation Answer: C) 2/63

Explanation:

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

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30 32051
Q:

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) 3/7 B) 3/4
C) 4/5 D) 5/7
 
Answer & Explanation Answer: A) 3/7

Explanation:

X can take 7 values.
To get |X|+2) take X={−1,0,1}

=> P(|X|<2) = Favourable CasesTotal Cases = 3/7

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31 13419
Q:

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is :

A) 2/9 B) 1/9
C) 8/9 D) 7/9
 
Answer & Explanation Answer: B) 1/9

Explanation:

One person can select one house out of 3= 3C1 ways =3.

 

Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9.

 

Therefore, probability that all thre apply for the same house is 1/9

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50 18477
Q:

Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?

A) 52/221 B) 55/190
C) 55/221 D) 19/221
 
Answer & Explanation Answer: C) 55/221

Explanation:

We have n(s) =52C2 52 = 52*51/2*1= 1326. 

Let A = event of getting both black cards 

     B = event of getting both queens 

A∩B = event of getting queen of black cards 

n(A) = 52*512*1 = 26C2 = 325, n(B)= 26*252*1= 4*3/2*1= 6  and  n(A∩B) = 4C2 = 1 

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and 

P(A∩B) = n(A∩B)/n(S) = 1/1326 

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

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359 113166
Q:

Two dice are thrown together .What is the probability that the sum of the number on the two faces is divided by 4 or 6.

A) 7/18 B) 14/35
C) 8/18 D) 7/35
 
Answer & Explanation Answer: A) 7/18

Explanation:

Clearly, n(S) = 6 x 6 = 36
Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.
Then,E = {(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6)}
n(E) = 14.
Hence, P(E) = n(E)/n(S) = 14/36 = 7/18

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105 37671
Q:

A bag contains 6 white and 4 black balls .2 balls are drawn at random. Find the probability that they are of same colour.

A) 1/2 B) 7/15
C) 8/15 D) 1/9
 
Answer & Explanation Answer: B) 7/15

Explanation:

Let S be the sample space

 

Then n(S) = no of ways of drawing 2 balls out of (6+4) =10C2 10 =10*92*1 =45

 

Let E = event of getting both balls of same colour

 

Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)

 

                =6C2+4C2 = 6*52*1+4*32*1= 15+6 = 21

 

Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15

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752 129785
Q:

In a simultaneous throw of pair of dice. Find the probability of getting the total more than 7.

A) 1/2 B) 5/12
C) 7/15 D) 3/12
 
Answer & Explanation Answer: B) 5/12

Explanation:

Here n(S) = (6 x 6) = 36

Let E = event of getting a total more than 7
        = {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}

Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12.

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