539
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:   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

Q:

A five digit number is formed with the digits 0,1,2,3 and 4 without repetition. Find the chance that the number is divisible by 5.

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

Explanation:
5 digit number = 5! = 120
Divisible by 5 then the last digit should be 0
Then the remaining position have the possibility = 4!
= 24
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5 198
Q:

In a single throw with 2 dices, what is probability of neither getting an even number on one and nor a multiple of 3 on other?

A) 11/36 B) 25/36
C) 5/6 D) 1/6
 
Answer & Explanation Answer: B) 25/36

Explanation:
We first calculate the probability of getting an even number on one and a multiple of 3 on other,Here, n(s) = 6x6 = 36 andE = (2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4)(3,6) (6,2)(6,4)n(E) = 11P(E) = 11/36Required probability = 1-11/36 = 25/36
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1 196
Q:

A bag contains 2 red Roses, 4 yellow Roses and 6 pink Roses. Two roses are drawn at random. What is the probability that they are not of same color?

A) 1/6 B) 2/3
C) 14/33 D) 5/6
 
Answer & Explanation Answer: B) 2/3

Explanation:

Possible outcomes = (RY, YP, PR)

2C1 4C1 + 4C1 6C1 + 6C1 2C1

Required probability = (2C1 4C1 + 4C1 6C1 + 6C1 2C1)/(12C2)

= 8 + 24 + 12/66

= 44/66

= 2/3.

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36 2404
Q:

What is the probability that a number is divisible by 4?

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

Explanation:

Probability is Number of favourable outcomes by total outcomes

Total comes = 2 (Divisible or not)

Favourable = 1 (Divisible)

 

Hence, probability = 1/2.

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23 1493
Q:

The probability that z lies between 0 and 3.01

A) 0.4987 B) 0.5
C) 0.9987 D) 0.1217
 
Answer & Explanation Answer: A) 0.4987

Explanation:

Probability between z = 0 and z = 3.01 is given by

P(0<z<3.01) = P(z<3.01) - P(z<0)

Reading from the z-table, we have

P(z<0) = 0.5

P(z<3.01) = 0.9987

Hence, P(0<z<3.01) = 0.9987 - 0.5 = 0.4987.

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22 2807
Q:

There are 26 balls marked with alphabetical order A to Z. What is the probability of selecting vowels listed balls? 

A) 1 B) 21/26
C) 5/26 D) 5
 
Answer & Explanation Answer: C) 5/26

Explanation:

We know that,

Total number of balls n(S) = 26

Number of vowels n(E) = 5

Hence, required probability = n(E)/n(S) = 5/26.

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24 1679
Q:

A group of five persons is formed from five boys and four girls. Find the probability that there are at least two girls in the group?

A) 110/129 B) 5/6
C) 121/126 D) 3/7
 
Answer & Explanation Answer: A) 110/129

Explanation:

Total number of possible ways = 

9C5 = 9 x 8 x 7 x 6 x 55 x 4 x 3 x 2 x 1 = 126

Number of favorable cases = 

4C2 x 5C3 + 4C3 x 5C2 + 4C4 x 5C14 x 32 x 1 x 5 x 4 x 33 x 2 x 1 + 4 x 3 x 23 x 2 x 1 x 5 x 4 2 x 1 + 1 x 5= 6 x 10 + 4 x 10 + 5= 60 + 40 + 5= 105

 

Therefore, required probability = 105/126 = 5/6

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6 1678
Q:

Out of 3 girls and 6 boys a group of three members is to be formed in such a way that at least one member is a girl. In how many different ways can it be done?

A) 64 B) 84
C) 56 D) 20
 
Answer & Explanation Answer: A) 64

Explanation:

Total number of possible ways = 9C3 = 84 ways

Required atleast one girl in the group of three = total possible ways - ways in which none is girl

None of the members in the group is girl = 6C3 = 20

 

Therefore, number of ways that at least one member is a girl in the group of three

= 84 - 20

= 64 ways.

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12 1708