6
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:

A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved?

A) 1/4 B) 1/2
C) 3/4 D) 7/12
 
Answer & Explanation Answer: C) 3/4

Explanation:

Let A, B, C be the respective events of solving the problem and A , B, C be the respective events of not solving the problem. Then A, B, C are independent event

A, B, C are independent events

Now,  P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

 PA=12, PB=23, PC= 34

 P( none  solves the problem) = P(not A) and (not B) and (not C)  

                  = PABC 

                  = PAPBPC          A, B, C are Independent                       

                  =  12×23×34  

                  = 14  

Hence, P(the problem will be solved) = 1 - P(none solves the problem) 

                = 1-14= 3/4

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Exam Prep: AIEEE , Bank Exams , CAT , GATE
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996 203380
Q:

An urn contains 4 white 6 black and 8 red balls . If 3 balls are drawn one by one without replacement, find the probability of getting all white balls.

A) 5/204 B) 1/204
C) 13/204 D) None of these
 
Answer & Explanation Answer: B) 1/204

Explanation:

Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then 

 Required probability = PABC 

PA PBA PCAB

 Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9

When  a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white

 PBA=317 

Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.

 PCAB =216=18

 Hence the required probability = 29×317×18=1204

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25 21943
Q:

A coin is tossed twice if the coin shows head it is tossed again but if it shows a tail then a die is tossed. If 8 possible outcomes are equally likely. Find the probability that the die shows a number greater than 4, if it is known that the first throw of the coin results in a tail

A) 1/3 B) 2/3
C) 2/5 D) 4/15
 
Answer & Explanation Answer: A) 1/3

Explanation:

Here Sample space S = { HH, HT, T1, T2, T3, T4, T5, T6 }

 

Let A be the event that the die shows a number greater than 4 and B be the event that the first throw of the coin results in a tail then,

 

 A = { T5, T6 }

 

 B = { T1, T2, T3, T4, T5, T6 }

 

Therefore, Required probability = PAB=PABPB=2868=13

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15 10300
Q:

A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?

A) 136/345 B) 17/87
C) 316/435 D) 158/435
 
Answer & Explanation Answer: C) 316/435

Explanation:

ns=C230

 

 Let A be the event of getting two oranges and 

 

 B be the event of getting two non-defective fruits.

 

 and AB be the event of getting two non-defective oranges

 

  PA=C220C230, PB=C222C230 and PAB=C215C230

 

 PAB=PA+PB-PAB

 

C220C230+C222C230-C215C230=316435

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107 46020
Q:

Two cards are drawn at random from a well - shuffled pack of 52 cards. what is the probability that either both are red or both are queens?

A) 17/112 B) 55/221
C) 55/121 D) 33/221
 
Answer & Explanation Answer: B) 55/221

Explanation:

n(S) = C252 = 1326

 

 Let  A = event of getting both red cards

 

and B = event of getting both queens

 

then AB = event of getting two red queens

 

n(A) = C226 = 325,   n(B) = C24 = 6

 

 n(AB)=C22=1

 

  PA=3251326, PB = 61326

 

 PAB=11326

 

P ( both red or both queens) = PAB

 

PA+PB-PAB=3251326+1221-11326=55221

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6 4020
Q:

In a single throw of two dice , find the probability that neither a doublet nor a total of 8 will appear.

A) 7/15 B) 5/18
C) 13/18 D) 3/16
 
Answer & Explanation Answer: B) 5/18

Explanation:

n(S) = 36

 

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

 

B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) }

 

nA=6, nB=5, nAB=1

 

Required probability = PAB

 

 = PA+PB-PAB

 

=  636+536-136 = 518

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43 20246
Q:

The probability of occurance of two two events A and B are 1/4 and 1/2 respectively. The probability of their simultaneous occurrance is 7/50. Find the probability that neither A nor B occurs.

A) 25/99 B) 39/100
C) 61/100 D) 17/100
 
Answer & Explanation Answer: B) 39/100

Explanation:

P ( neither A nor B) = PA and B  

 

=  PAB=  PAB1-PAB  

 

1-61100=39100

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8 9213
Q:

A box contains 5 detective and 15 non-detective bulbs. Two bulbs are chosen at random. Find the probability that both the bulbs are non-defective.

A) 5/19 B) 3/20
C) 21/38 D) None of these
 
Answer & Explanation Answer: C) 21/38

Explanation:

n(S) = C220 = 190 

n(E) = C215 = 105 

Therefore, P(E) = 105/190 = 21/38

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11 14618