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# Probability Questions for SNAP

Probability is an important topic in the Quant section of the SNAP Exam. Quant is a scoring section in SNAP, so it is advised to practice as much as questions from quant. This article provides some of the most important Probability Questions for SNAP. One can also download this Free Probability Questions for SNAP PDF with detailed answers by Cracku. These questions will help you practice and solve the Probability questions in the SNAP exam. Utilize this PDF practice set, which is one of the best sources for practicing.

Question 1:Â The arithmetic mean of all the distinct numbers that can be obtained by rearrangingÂ the digits in 1421, including itself, is

a)Â 2222

b)Â 2442

c)Â 2592

d)Â 3333

Solution:

The number of 4-digit numbers possible using 1,1,2, and 4 isÂ $\frac{4!}{2!}=12$

Number ofÂ 1’s, 2’s and 4’s in units digits will be in the ratio 2:1:1, i.e. 6 1’s, 3 2’s and 3 4’s.

Sum = 6(1) + 3(2) + 3(4) = 24

Similarly, in tens digit, hundreds digit and thousands digit as well.

Therefore, sum = 24 + 24(10) + 24(100) + 24(1000) = 24(1111)

Mean =Â $\frac{24\left(1111\right)}{12}=2222$

Question 2:Â The number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once, is

a)Â 1440

b)Â 1200

c)Â 1480

d)Â 1420

Solution:

Case 1: 4-digit numbers

Given digits – 0, 1, 2, 3, 4, 5

_, _, _, _

As the numbers should be greater than 2000, first digit can be 2, 3, 4 and 5.

For remaining digits, we need to arrange 3 digits from the remaining 5 digits, i.e. 5*4*3 = 60 ways

Total number of possible 4-digit numbers = 4*60 = 240

Case 2: 5-digit numbers

_, _, _, _, _

First digit cannot be zero.

Therefore, total number of cases = 5*5*4*3*2 = 600

Case 3: 6-digit numbers

_, _, _, _, _, _

First digit cannot be zero.

Therefore, total number of cases = 5*5*4*3*2*1 = 600

Total number of integers possible = 600 + 600 + 240 = 1440

Question 3:Â The number of ways of distributing 20 identical balloons among 4 children such that each child gets some balloons but no child gets an odd number of balloons, is

Solution:

Let the number of balloons each child received be 2a, 2b, 2c and 2d

2a + 2b + 2c + 2d = 20

a + b + c + d = 10

Each of them should get more than zero balloons.

Therefore, total number of ways =Â $(n-1)_{C_{r-1}}=(10-1)_{C_{4-1}}=9_{C_3}=84$

Question 4:Â A bag contains 6 red balls, 11 yellow balls and 5 pink balls. If two balls are drawn at a random from the bag, one after another. What is the probability that the first ball drawn was red and the second ball drawn was yellow in colour?

a)Â $\frac{3}{14}$

b)Â $\frac{5}{7}$

c)Â $\frac{1}{7}$

d)Â $\frac{1}{14}$

Solution:

There are 6 red balls, 11 yellow balls and 5 pink balls

Total number of balls = 6 + 11 + 5 = 22

Probability of selecting red ballÂ =Â $\frac{6}{22}$

After selecting one ball, there are 21 balls left

Probability of selecting yellow ballÂ = $\frac{11}{21}$

Therefore, probability =Â $\frac{6}{22}\times\ \frac{11}{21}$ =Â $\frac{1}{7}$

Question 5:Â In a bag there are 15 red balls and 10 green balls. Three balls are selected at random. The probability of selecting 2 red balls and 1 green ball is :

a)Â $\frac{21}{46}$

b)Â $\frac{25}{117}$

c)Â $\frac{3}{25}$

d)Â $\frac{1}{50}$

Solution:

Number of ways of selecting 2 red balls and 1 green ball =Â $15_{C_2}.\ 10_{C_1}$

Number of ways of selecting 3 balls =Â $25_{C_3}$

Probability of selecting 2 red balls and 1 green ball =Â $\ \frac{15_{C_2}.10_{C_1}}{25_{C_3}}$ =Â $\frac{21}{46}$

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Question 6:Â The following five candidates have applied for a position in an organisation: Male 30Â years, Male 32 years, Female 45 years, Female 20 years, and Male 40 years. What is theÂ probability that the candidate selected for the position will be either female or over 35Â years?

a)Â 4/5

b)Â 2/5

c)Â 1/5

d)Â 3/5

Solution:

Female – A
above 35 years – B
n(A $\cup$ B ) = n(A)+n(B) – n (A $\cap$ B) = 2 + 2 – 1 = 3
Probability =Â $\frac{3}{5}$

Question 7:Â In a class of 45 students, 25 are girls. In a test, 30 students scored above 90% marksÂ and 18 of them are girl students. A student is selected at random. The probability ofÂ selecting a girl scoring above 90% marks is___________

a)Â 18/45

b)Â 12/45

c)Â 18/30

d)Â 30/45

Solution:

It is given,
Total number of students = 45
Number of girls who scored above 90% marks = 18
Therefore, probability =Â $\frac{18}{45}\$

Question 8:Â In a charity show, tickets numbered consecutively from 101 through 350 are placed inÂ a box. What is the probability that a ticket selected randomly will have a number with aÂ hundredth digit as two?

a)Â 100/250

b)Â 99/249

c)Â 100/249

d)Â 99/250

Solution:

350 = 101 + (n-1)1
n = 250
Total number of tickets = 250
There are 100 tickets(200 to 299) between the tickets 101 and 350 whose hundredth digit is 2.
Probability =Â $\ \frac{100}{250}$

Question 9:Â What is the probability that a leap year has 53 Sundays and 53 Mondays?

a)Â 2/7

b)Â 1

c)Â 0

d)Â 1/7

Solution:

Number of days in leap year = 366
There are 2 odd days in a leap year. For a year to have 53 sundays and 53 mondays, 2 odd days should be sunday and monday.
This is only possible when the first day of the year is sunday.
Therefore, the probability of the first day is sunday = $\frac{1}{7}$

Question 10:Â There is a shooting event organised by the SBT youth Club to select the best candidate who will qualify to participate in Eklavya championship. The shooting board is designed by using 4 concentric circles of radii 2 inch, 3 inch, 5 inch and 9 inch. What is the probability that the participant will shoot only in the second ring [from the outside]Â to qualify for Eklavya championship.

a)Â $\frac{16}{25}$

b)Â $\frac{20}{81}$

c)Â $\frac{20}{56}$

d)Â $\frac{16}{81}$

Solution:

The board is as follows.

The required probability =Â $\frac{\pi\left(5\right)^2-\pi\left(3\right)^2}{\pi\left(9\right)^2}=\frac{25-9}{81}=\frac{16}{81}$

Hence, the answer is option D.

Question 11:Â How many 4 letter words can be formed from the word “CORONAVIRUS”.

a)Â 3148

b)Â 3058

c)Â 3702

d)Â 3086

Solution:

“CORONAVIRUS” has 7 distinct alphabets and two pairs of repeated characters “, O” and “R”.

There are three possible cases for creating 4 letter words.

1. Two letters are “O” and the other two are “R”.

The total number of arrangements =Â $\frac{4!}{2!\times2!}=6$

2. Two of the letters are either “O” or “R” and the others are distinct.

The total number of arrangements =Â $^2C_1\times^8C_2\times\frac{4!}{2!}=2\times28\times12=672$

3. All four letters are distinct.

The total number of arrangements =Â $^9C_4\times4!=3024$

Thus, the total number of four-letter words possible = $6+672+3024=3702$.

Hence, the answer is option C.

Question 12:Â Care hospital has vaccination facility at different floors of the hospital i.e. Covishield at first floor, Covaxin at 2nd floor and Sputnik at 3rd floor. Daily quota of allotted vaccine is determined by the government. On a particular day, 35% of the patients got Covaxin, 45% of the patients got Covishield and 20% got Sputnik. After vaccination, the vaccinated people were instructed to take Paracetamol if they experienced fever. Among the vaccinated people, those who received Covaxin have 15% chance of getting fever, Covishield vaccinated people have 10% chance of getting fever, and Sputnik vaccinated people have 5% chance of getting fever. A vaccinated person was randomly picked up and it was found that he had a running temperature at 102F. What is the chance that he had been administered Covishield vaccine?

a)Â 49%

b)Â 42%

c)Â 58%

d)Â 45%

Solution:

Required probability =Â $\ \ \frac{\ \left(0.45\right)\left(0.1\right)}{\left(0.35\right)\left(0.15\right)+\left(0.45\right)\left(0.1\right)+\left(0.2\right)\left(0.05\right)}\times\ 100$ =Â $\ \ \frac{0.045}{0.1075}\times\ 100$ = 41.86%Â $\approx\$ 42%

Question 13:Â Shanu and Ishu play a tossing game. They alternatively toss a coin. Whoever receives ‘Head’ first, wins the game. What is Ishu’s chance of winning the game if Shanu gets a chance to toss coin first?

a)Â $\frac{1}{3}$

b)Â $\frac{2}{3}$

c)Â $\frac{1}{2}$

d)Â $\frac{1}{4}$

Solution:

Probability ofÂ Ishu winning in first round = P(Shanu tail)P(Ishu head) =Â $\frac{1}{2}\cdot\frac{1}{2}$ =Â $\left(\frac{1}{2}\right)^2$

Probability of Ishu winning in second round = P(Shanu tail)P(Ishu tail)P(Shanu tail)P(Ishu head) =Â $\left(\frac{1}{2}\right)^4$

Similarly, calculating for further rounds we get

Probability =Â $\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^{6\ }+\ …..\ =\ \frac{1}{3}$

Question 14:Â An unbiased dice is tossed seven times. Find the probability of getting a third six on the seventh throw.

a)Â $\frac{\left(\begin{array}{c}6\\ 2\end{array}\right)5^2}{6^7}$

b)Â $\frac{\left(\begin{array}{c}6\\ 2\end{array}\right)5^4}{6^7}$

c)Â $\frac{\left(\begin{array}{c}6\\ 3\end{array}\right)5^3}{6^7}$

d)Â $\frac{\left(\begin{array}{c}7\\ 3\end{array}\right)5^4}{6^7}$

Solution:

We need a third 6 in the 7th throw which means that in the first 6 throws we should get 6 exactly twice.

No. of ways of getting 6 exactly twice in first 6 throws=Â $^6C_2$

Probability of getting 6 for the third time in the seventh throw=Â $\frac{\left(^6C_2\times\ 5^4\right)}{6^7}$

Question 15:Â Ishika speaks truth in 60% of cases and Mishika in 85% of cases. Ishika and Mishika agree in astatement. Find the probability that the statement is true.

a)Â $\frac{49}{57}$

b)Â $\frac{51}{100}$

c)Â $\frac{57}{100}$

d)Â $\frac{51}{57}$

Solution:

For both Ishika and Mishika to agree on a statement either both have to speak the truth or both have to speak false.

Probability of both speaking truth = (60/100)*(85/100) = 204/400

Probability of both speaking false = (40/100)*(15/100) = 24/400

The probability that the statement is true when Ishika and Mishika both agree on a statement = (204/400)/{(204/400)+(24/400)} = 204/228 = 51/57

Question 16:Â A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is

Solution:

The question asks for the number of 4 digit numbers using only the digits 1, 2, and 3 such that the digits 2 and 3 appear at least once.

The different possibilities include :

Case 1:The four digits are ( 2, 2, 2, 3). Since the number 2 is repeated 3 times. The total number of arrangements are :

$\frac{4!}{3!}$ = 4.

Case 2: The four digits are 2, 2, 3, 3. The total number of four-digit numbers formed using this are :

$\frac{4!}{2!\cdot2!}=\ 6$

Case 3: The four digits are 2, 3, 3, 3. The number of possible 4 digit numbers are :

$\frac{4!}{3!}$ = 4

Case4: The four digits are 2, 3, 3, 1. The number of possible 4 digit numbers are :

$\frac{4!}{2!}=\ 12$

Case5: Using the digits 2, 2, 3, 1. The number of possible 4 digit numbers are :

$\frac{4!}{2!}=\ 12$

Case 6: Using the digits 2, 3, 1, 1. The number of possible 4 digit numbers are :

$\frac{4!}{2!}=\ 12$

A total of 12 + 12 + 12 + 4 + 6 + 4 = 50 possibilities.

Alternatively

We have to form 4 digit numbers using 1,2,3 such that 2,3 appears at least once
So the possible cases :

Now we getÂ $\frac{4!}{2!}\times\ 3$= 36 ( When one digit is used twice and the remaining two once )
$\frac{4!}{3!}\times\ 2$ = 8 ( When 1 is used 0 times and 2 and 3 is used 3 times or 1 time )
$\frac{4!}{2!\times\ 2!}=\ 6$( When 2 and 3 is used 2 times each )
So total numbers = 36+8+6 =50

Question 17:Â The number of ways of distributing 15 identical balloons, 6 identical pencils and 3 identical erasers among 3 children, such that each child gets at least four balloons and one pencil, is

Solution:

This question is an application of the product rule in probability and combinatorics.

In the product rule, if two events A and B can occur in x and y ways, and for an event E, both events A and B need to take place, the number of ways that E can occur is xy. This can be expanded to 3 or more events as well.

Event 1: Distribution of balloons

Since each child gets at least 4 balloons, we will initially allocate these 4 balloons to each of them.

So we are left with 15 – 4 x 3 = 15 – 12 = 3 balloons and 3 children.

Now we need to distribute 3 identical balloons to 3 children.

This can be done inÂ $^{n+r-1}C_{r-1}$ ways, where n = 3 and r = 3.

So, number of ways =Â $^{3+3-1}C_{3-1}=^5C_2=\frac{5\times\ 4}{2\times\ 1}=10$

Event 2: Distribution of pencils

Since each child gets at least one pencil, we will allocate 1 pencil to each child. We are now left with 6 – 3 = 3 pencils.

We now need to distribute 3 identical pencils to 3 children.

This can be done in $^{n+r-1}C_{r-1}$ ways, where n = 3 and r = 3.

So, number of ways = $^{3+3-1}C_{3-1}=^5C_2=\frac{5\times\ 4}{2\times\ 1}=10$

Event 3: Distribution of erasers

We need to distribute 3 identical erasers to 3 children.

This can be done in $^{n+r-1}C_{r-1}$ ways, where n = 3 and r = 3.

So, number of ways = $^{3+3-1}C_{3-1}=^5C_2=\frac{5\times\ 4}{2\times\ 1}=10$

Applying the product rule, we get the total number of ways = 10 x 10 x 10 = 1000.

Question 18:Â How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?

Solution:

Let the numbers be of the form 100a+10b+c, where a, b, and c represent single digits.

Then (100c+10b+a)-(100a+10b+c)=198

99c-99a=198

c-a = 2.

Now, a can take the values 1-7. a cannot be zero as the initial number has 3 digits and cannot be 8 or 9 as then c would not be a single-digit number.

Thus, there can be 7 cases.

B can take the value of any digit from 0-9, as it does not affect the answer. Hence, the total cases will beÂ $7\times\ 10=70$.

Question 19:Â The number of groups of three or more distinct numbers that can be chosen from 1, 2, 3, 4, 5, 6, 7 and 8 so that the groups always include 3 and 5, while 7 and 8 are never included together is

Solution:

The possible arrangements are of the form

35 _ Can be chosen in 6 ways.

35 _ _Â We can choose 2 out of the remaining 6 inÂ $^6C_2=15$ways. We remove 1 case where 7 and 8 are together to get 14 ways.

35 _ _ _We can choose 3 out of the remaining 6 in $^6C_3=20$ways. We remove 4 cases where 7 and 8 are together to get 16 ways.

35 _ _ _ _We can choose 4 out of the remaining 6 in $^6C_4=15$ways. We remove 6 case where 7 and 8 are together to get 9 ways.

35 _ _ _ _ _ We choose 1 out of 7 and 8 and all the remaining others in 2 ways.

Thus, total number of cases = 6+14+16+9+2 = 47.

Alternatively,

The arrangement requires a selection of 3 or more numbers while including 3 and 5 and 7, 8 are never included together. We have cases including a selection of only 7, only 8 and neither 7 nor 8.

Considering the cases, only 7 is selected.

We can select a maximum of 7 digit numbers. We must select 3, 5, and 7.

Hence we must have ( 3, 5, 7) for the remaining 4 numbers we have

Each of the numbers can either be selected or not selected and we have 4 numbers :

Hence we have _ _ _ _ and 2 possibilities for each and hence a total of 2*2*2*2 = 16 possibilities.

SImilarly, including only 8, we have 16 more possibilities.

Cases including neither 7 nor 8.

We must have 3 and 5 in the group but there must be no 7 and 8 in the group.

Hence we have 3 5 _ _ _ _.

For the 4 blanks, we can have 2 possibilities for either placing a number or not among 1, 2, 4, 6.

= 16 possibilities

But we must remove the case where neither of the 4 numbers are placed because the number becomes a two-digit number.

Hence 16 – 1 = 15 cases.

Total = 16+15+16 = 47 possibilities

Question 20:Â An article manufactured by a company consists of two parts: Part I and Part II. In theprocess of manufacturing of Part I, 9 products out of 100 are likely to be defective.Similarly, 5 products out of 100 are likely to be defective in the manufacture of Part II.What is the probability that the final assembled article will NOT be defective?

a)Â 0.8465

b)Â 0.8645

c)Â 0.8564

d)Â 0.8546

Probability to be defective for part I = $\frac{9}{100}$
Probability to be defective for part II =Â $\frac{5}{100}$
Probability =Â $\frac{91}{100}\times\ \frac{95}{100}$ = 0.8645