6
Q:

Four dice are thrown simultaneously. Find the probability that two of them show the same face and remaining two show the different faces.

A) 4/9 B) 5/9
C) 11/18 D) 7/9

Answer:   B) 5/9



Explanation:

Select a number which ocurs on two dice out of six numbers (1, 2, 3, 4, 5, 6). This can be done in C16, ways.

 

Now select two distinct number out of remaining 5 numbers which can be done in C25 ways. Thus these 4 numbers can be arranged in 4!/2! ways.

 

So, the number of ways in which two dice show the same face and the remaining two show different faces is 

 C16×C25×4!2!=720

 =>  n(E) = 720

 PE=72064=59

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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6 344
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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2 311
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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8 612
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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14 930
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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3 768
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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7 751
Q:

A bag contains 5 red smileys, 6 yellow smileys and 3 green smileys. If two smileys are picked at random, what is the probability that both are red or both are green in colour?

A) 7 B) 1/7
C) 3/7 D) 0
 
Answer & Explanation Answer: B) 1/7

Explanation:

Given total number of smileys = 5 + 6 + 3 = 14

Now, required probability = 

5C214C2 + 3C214C2 = 1091 + 391 = 1391 = 17

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3 861
Q:

When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 12?

A) 35/36 B) 17/36
C) 15/36 D) 1/36
 
Answer & Explanation Answer: A) 35/36

Explanation:

When two dice are thrown simultaneously, the probability is n(S) = 6x6 = 36

dice_thrown_simulataneously1532668754.png image

Required, the sum of the two numbers that turn up is less than 12

That can be done as n(E)

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

= 35

Hence, required probability = n(E)/n(S) = 35/36.

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9 1247