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# Probability Questions

Home > Quantitative Aptitude > Probability > General Questions
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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:

A21/46

B25/117

CJan-50

D25-Mar

ENone of these

Explanation:

Let's assume the sample space = S
and Event of selecting 1 girl and 2 boys = E

So, n(S) = Number ways of selecting 3 students out of 25 = 25C3
=> (25 * 24 * 23)/(3 * 2 * 1) = 2300.
n(E) = (10C1 * 15C2)
= 10 * [(15 * 14)/(2 * 1)] = 1050.
P(E) = n(E)/n(S) = 1050/2300 = 21/46.

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In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

A1/10

B2/5

C2/7

D5/7

Explanation:

P (getting a prize) = 10/(10 + 25)
=> 10/35 = 2/7

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A 5-digit number is formed by the digits 2,4,5,6,8 (each digit used exactly once) . What is the probability that the number formed is divisible by 4 ?

A1/5

B2

C3

D2/5

Explanation:

A no. is divisible by 4 if last two digit is divisible by 4
so numbers ending with 24,28,48,52,56,64,68,84 are divisible by 4
last two places are fixed by 24,28,48,52,56,64,68,84
so remaining 3 places can be filled in 3! ways for each
total=8*3!=48=n(E)
n(S) = 5!
so
= n(E)/n(S)
=48/120=2/5

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A group consists of equal number of men and women. Of them 10% of men and 45% of women are unemployed. If a person is randomly selected from the group. Find the probability for the selected person to be an employee.

A15/40

B30/40

C29/40

D19/40

Explanation:

Lets assume men=100, women=100 then employed men & women = (100-10)+(100-45) = 145
So probability for the selected person to be an employee = (145/200) = 29/40

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Aniket and kumari are a married couple. They have two kids, one of them is a girl. Assume safely that the probability of each gender is 1/2.What is the probability that the other kid is also a girl?

A1/2

B1/3

C2/3

DNone of these

Explanation:

There are 4 possible conditions of two children.

Girl â€“ Girl, Girl â€“ Boy, Boy â€“ Girl, Boy - Boy

Since we know one of the children is a girl, we will drop the Boy-Boy possibility from the sample space.

This leaves only three possibilities, one of which is two girls. Hence the probability is 1/3

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A drawer hold 4 red hates and 4 blue hates. what is the probability of getting exactly 3 hatesv when taking out 4 hates randomly out of the drawerband imediately returining every hate to the drawer before taking out the next?

A8-Mar

B8-May

C8-Jul

D8-Jan

Explanation:

The probability of drawing red or blue is equal.
Probability to draw exactly 3 red hats and 1 blue hat =1/2*1/2*1/2*1/2=1/16
Similarly probability to draw exactly 3 blue hats and 1 red hat =1/2*1/2*1/2*1/2=1/16
Total probability =1/16+1/16=1/8

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A man speaks truth 3 out of 4 times. He throws a die and reports it to be a 6.What is the probability of it being a 6?

A8-Jul

B8-Jan

C8-Mar

D8-May

Explanation:

Probability that the die does not report 6, and he is telling the truth: (5/6)*(3/4)=5/8
Therefore, Probability that the die reports 6 = 1-(5/8) = 3/8

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2 containers labeled A and B are filled with red and blue marbles in given quantities: -
Container RED BLUE
A 6 4
B 60 40
Each container is shaken vigorously. After choosing 1 of the container, without looking, draw out a marble. From which container you have the max probability of choosing a blue marble?

AContainer A (6 red, 4 blue)

BContainer B (60 red, 40 blue)

CEqual chances from each container

DNone of these

Explanation:

If any one choose container A then prob = (1/2)*(4/10)=1/5
If any one choose container A then prob = (1/2)*(40/100)=1/5

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From a railway station, trains leave for every 15 minutes and 25 minutes to city A and city B respectively. First train to city A and city B start at 9 am and 10:15 am respectively. If a man arrives to the station in between 11:25 am and 12:25 pm then the probability of getting train for city A is:

A1/4

B4/7

C3/7

D2/5

Explanation:

Given man arrived station in between 11.25 am and 12.25 pm.
First train to city A is at 9 am and there is a train for every 15 minutes.
Trains for city A will leave at the following times : 9:00 am, 9:15 am, 9:30 am,.......,11:30 am, 11:45 am, 12:00 pm, 12:15pm, and so on.
Number of trains for city A between 11:25 am and 12:25 pm is 4.
First train to city B is at 10:15 am and there is a train for every 25 minutes.
Trains for city B will leave at the following times: 10:15 am, 10:40 am, 11:05 am, 11:30 am, 11:55 am, 12:20 pm, and so on.
Number of trains for city B between 11:25 am and 12:25 pm is 3.
Probability of getting train for city A between 11:25 am and 12:25 pm = Number trains for city A from 11:25 am to 12:25 pm / Total number of trains for city A and B from 11:25 am to 12:25 pm
= 4/7

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There are two bus stands, namely X and Y. Buses leave from X for every 30 minutes and its first bus starts at 8:05 am. Every hour number of buses leaving from Y increases by 1 and its first bus starts at 7:00 am. From Y there is only 1 bus for the 1st hour. Any bus from either of the bus stations takes 15 minutes to reach a nearby bus stop. Suppose a person reaches the stop in between 12:15 pm and 1:15 pm. The probability that the person will get a bus from Y is:

A1/4

B1

C2/3

D3/4

Explanation:

From bus stand X :
The first bus will leave by 8.05 am and reach the bus stop in 15 minutes, i.e. at 8.20 am
The second bus will leave after 30 minutes i.e. at 8.35 am and will reach the stop at 8.50 am
Therefore, buses will reach the stop at the following times: 8.20am, 8.50am, 9.20 am,.....,12.20 pm, 12.50 pm, 1.20 pm and so on.
Between 12.15 pm and 1.15 pm, two buses will reach the stop at 12.20 pm and 1.20 pm.
Therefore, the person will get 2 buses from X.
From bus stand Y :
The first bus will leave by 7 am and reach the bus stop in 15 minutes, i.e. at 7.15 am.
There is only one bus for first 1 hour.
i.e., the second bus will leave at 8 am.
Note that, the number of buses leaving from Y is increased by 1 per hour.
From 8 am to 9 am, two buses will leave from Y and reach the stop between 8.15 am to 9.15 am.
And from 9 am to 10 am, 3 buses will leave from Y and reach the stop between 9.15 am to 10.15 am.
Proceeding like this, we have,
From 12 pm to 1 pm, 6 buses will leave from Y and reach the stop between 12.15 pm to 1.15 pm.
Therefore, the person will get 6 buses from Y between 12.15 pm to 1.15 pm.
The probability of getting the bus from Y between 12.15 pm to 1.15 pm = Number buses from Y in between 12.15 pm to 1.15 pm / Total number of buses from X and Y in between 12.15 pm to 1.15pm = 6/(2+6) = 6/8 = 3/4.

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