Directions: Study the given information carefully and answer the questions that follow—

A store contains 5 red, 4 blue, 5 green shirts.

If two shirts are picked at random, what is the probability that at most one is blue?

Probabilities if at most one is blue =

[(^{4}C_{0}*^{10}C_{2}+ ^{4}C_{1}*^{10}C_{1})/^{14}C_{2}] = (1*45 + 4*10)/( 91) = 85/91

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Direction: Study the following information carefully and answer the questions that follow:

A committee of 10 persons is to be formed from 7 men and 6 women.

In how many ways this can be done if at least 5 men to have to be included in a committee.

Number of Ways when if at least 5 men include in committee = ^{7}C_{5}*^{6}C_{5}+^{7}C_{6}*^{6}C_{4}+^{7}C_{7}*^{6}C_{3}

= 21*6+7*15+1*20

= 251

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Direction: Study the following information carefully and answer the questions that follow:

A committee of 10 persons is to be formed from 7 men and 6 women.

In how many ways of these committee the women are in majority

Number of ways when women are majority in committee =

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Direction: Study the following information carefully and answer the questions that follow:

A committee of 10 persons is to be formed from 7 men and 6 women.

In how many of these committee the men are in majority

Number of ways when men are majority in committee = ^{7}C_{6}*^{6}C_{4}+^{7}C_{7}*^{6}C_{3}

7*15+1*20 = 125

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Direction: Study the following information carefully and answer the questions that follow:

A committee of 10 persons is to be formed from 7 men and 6 women.

In how many ways this can be done if 6 men and 4 women be included in a committee

If 6 men and 4 women include in committee =

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Direction: Study the following information carefully and answer the questions that follow:

A committee of 10 persons is to be formed from 7 men and 6 women.

In how many ways this can be done if 5 men and 5 women be included in a committee

If 5 men and 5 women include in committee =

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In a basket there are 7 apples and 8 oranges. 4 fruits are picked at random. What is the probability that two fruits are apples and 2 are oranges?

**A. ** 72/455

**B. ** 82/455

**C. ** 89/455

**D. ** 84/455

**E. ** None of these

**Answer : ****Option E**

**Explaination / Solution: **

Total number of fruits n(s) =7 + 8=15

Probability = (^{7}C_{2}*^{8}C_{2})/^{15}C_{4}

= (21*28)/1365 = 28/65

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In how many ways the letter of the word IMMEDIATELY can be arranged so that vowels always come together ?

**A. ** 56800

**B. ** 75600

**C. ** 64800

**D. ** 84560

**E. ** None of these

**Answer : ****Option C**

**Explaination / Solution: **

There are 11 letters .out of which 5 are vowels and 6 are consonants.

So taking vowels together as a single letter , we have 7 letters

So no of arrangements =( 7! * 5!)/ (2!)^{3} [there are 2 i,m e]

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In how many ways can the word ENGINEER be arranged so that ‘G’ and ‘R’ are never together?

**A. ** 3120

**B. ** 2830

**C. ** 2520

**D. ** 3220

**E. ** None of these

**Answer : ****Option C**

**Explaination / Solution: **

Number of ways of rearranging the word ENGINEER = (8!)/(3!*2!) = 3360

Finding the number of ways of arranging the word ENGINEER such that G and R are always together is done by taking GR as a single alphabet and then finding the permutation.

Number of ways of arranging the word ENGINEER such that G and R are always together = (7!)/(3!*2!) = 420*2 = 840

∴Number of ways of arranging the word ENGINEER such that G and R are never together = Number of ways of rearranging the word ENGINEER - Number of ways of arranging the word ENGINEER such that G and R are always together

⇒Number of ways of arranging the word ENGINEER such that G and R are never together

= 3360 – 840

= 2520

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Read the following information to answer the following questions :

Two unbiased dice are thrown simultaneously

What is the probability of getting a doublet?

**A. ** 1/3

**B. ** 1/6

**C. ** 1/4

**D. ** 2/3

**E. ** None of these

**Answer : ****Option B**

**Explaination / Solution: **

Two unbiased dice are thrown simultaneously

What is the probability of getting a doublet?

Total possible outcomes = 6 × 6 = 36

E = Events of getting a doublet

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

PE = 6/36 = 1/6

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