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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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Q:

Find the probability of selecting 2 woman when four persons are choosen at random from a group of 3 men, 2 woman and 4 children.

A) 1/5	B) 1/7
C) 1/6	D) 1/9

Answer & Explanation Answer: C) 1/6

Explanation:

Out of 9 persons,4 can be choosen in $9 C_{4}$ ways =126.

Favourable events for given condition = $2 C_{2} * 7 C_{2}$ = 21.

So,required probability = 21/126 =1/6.

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Q:

A word consists of 9 letters; 5 consonants and 4 vowels.Three letters are choosen at random. What is the probability that more than one vowel will be selected ?

A) 13/42	B) 17/42
C) 5/42	D) 3/14

Answer & Explanation Answer: B) 17/42

Explanation:

3 letters can be choosen out of 9 letters in $9 C_{3}$ ways.

More than one vowels ( 2 vowels + 1 consonant or 3 vowels ) can be choosen in $(4 C_{2} * 5 C_{1}) + 4 C_{3}$ ways

Hence,required probability = $\frac{(4 C_{2} * 5 C_{1}) + 4 C_{3}}{9 C_{3}}$ = 17/42

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Q:

Find the probability of selecting exactly 2 children when four persons are choosen at random from a group of 3 men, 2 woman and 4 children.

A) 9/29	B) 10/21
C) 12/21	D) 14/19

Answer & Explanation Answer: B) 10/21

Explanation:

4 persons can be selected from 9 in $9 C_{4}$ ways =126

Fvaourable events = $4 C_{2} * 5 C_{2}$ =60

So,required probability = 60/126 = 10/21

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Q:

Four persons are to be choosen from a group of 3 men, 2 women and 4 children. Find the probability of selecting 1 man,1 woman and 2 children.

A) 2/7	B) 3/7
C) 4/7	D) 3/7

Answer & Explanation Answer: A) 2/7

Explanation:

Total number of persons = 9

Out of 9 persons 4 persons can be selected in $9 C_{4}$ ways =126

1 man,1 woman and 2 children can be selected in $3 C_{1} * 2 C_{1} * 4 C_{1}$ ways =36

So,required probability = 36/126 =2/7

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Q:

What is the probability that a leap year selected at random, will contain 53 sundays?

A) 1/7	B) 1/3
C) 2/7	D) 4/7

Answer & Explanation Answer: C) 2/7

Explanation:

In a leap year,there are 366 days=52 weeks and 2 days

Remaining favourable 2 days can be sunday and monday or saturday and sunday

Exhaustive number of cases =7

Favourable number of cases =2

So,required probability=2/7

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Q:

A number is selected from the numbers 1,2,3,4.......25.The probability for it to be divisible by 4 or 7 is:

A) 3/25	B) 9/25
C) 1/25	D) None of these

Answer & Explanation Answer: B) 9/25

Explanation:

Total numbers = 25

Numbers divisible by 4 or 7 are 4, 7, 8, 12, 14, 16, 20, 21, 24 = 9

The probability (divisible by 4 or 7) = 9/25

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Q:

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

A) 21/46	B) 1/5
C) 3/25	D) 1/50

Answer & Explanation Answer: A) 21/46

Explanation:

Let , S - sample space E - event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25

= $25 C_{3}$

= 2300.

n(E) = $10 C_{1} \times 15 C_{2}$ = 1050.

$\therefore$ P(E) = n(E)/n(s) = 1050/2300 = 21/46

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