 # Probability Solutions

Solutions for probability questions are given here. The solutions are explanatory, detailed and easily understandable to help the candidates muster the topic effectively.

### Test Your Probability Skills – Solutions

1) Option (d)

Probability of Getting an A – 10/50 = 1/5

Probability of getting a D – 30/49

Probability of getting a D – 29/48

Probability of getting a S – 10/47

Total probability= 1/5×30/49×29/48×10/47.

2) Option (c)

Probability of seating at least 2 children = 1- (Probability that the table seats none or exactly one child)

5 people can be selected out of 10 in 10C5 ways = 252

The number of ways of seating a table of 5 with 0 children= 0 (only 4 adults) Number of ways of seating a table with exactly one child= 6C1 =6 ways Required probability = 1 – 6/252 = 41/42.

3) Option (c)

If A is selected, then two more students need to be selected. This can be done in 5C2 ways= 10 If C is also selected, only one more out of the remaining 4 needs to be selected. That can be done in 4 ways. Probability = 4/10 = 0.4.

4) Option (c)

Probability of a success= 0.3 Probability of a failure= 0.7 Required Probability = 0.3 × 0.3 × 0.7 × 3C1 = 0.189.

5) Option (b)

Total sample space is 30 The number of multiples of 3 are 10 and the number of multiples of 13 are 2, out of which none of them are common. Hence, the answer is 2/5.

6) Option (d)

The probability of getting no six is (5/6)3 = 125/216. Hence the probability of getting at least one six is 1 – 125/216 = 91/216.

7) Option (d)

Sample space is 90 numbers, of which , 90/3 = 30 are multiples of 3. 90/5 = 18 are multiples of 5, of which every 1 out of 3 are multiples of both 3 and 5. Probability that a two-digit number is a multiple of 3 and not a multiple of 5 = 30 – 6 = 24/90 = 4/15.

8) Option b

First find the probability of sum being greater than 10.. Combination whose sum of 12 is (6, 6) Combinations whose sum of 11 is (5, 6), (6, 5). Therefore, there are totally 3 occurrences out of 36 occurrences that go against the given condition. Probability whose sum of two numbers is greater than or equal to 11 = 3/36 = 1/12. Hence probability whose sum of two numbers is lesser than 11 = 1 – 1/12 = 11/12. Option (b) is the answer.

9) Option (a)

Lets consider upto the 9th apple. The first 8 apples should have 3 rotten ones and remaining 5 good ones. This can be chosen in 4C3×11C5 ways. Total number of selections of 8 apples out of 15 apples is 15C8. The last apple is the only rotten one left, which can be selected in 1 way. Total number of ways of selecting that 1 apple from remaining 15 – 8 = 7 apples = 7C1 ways Total probability = 4C3×11C5/15C8×1/7C1

10) Option (c)

If he picks any of the three balls, invariably any two of them should be of the same colour. Hence the probability is 1.

11) Option (c)

The total possibilities are 2 × 2 × 2 = 8 (each customer has two possibilities). These are AAA, AAB, ABA, BAA, BBA, BAB, ABB, BBB. The favourable outcomes are only 4 (AAA, AAB, ABA, BAA). Thus the probability = 4/8 = 1/2.

12) Option (d)

No matter what sign Arpit throws, there is one sign Rita could throw that would beat it, one that would tie, and one that would lose. Rita is equally likely to throw any one of the three signs.

Therefore, the Probability that Arpit will win is 1/3.

Probability that Rita will win = 1/3

Probability of a tie = 1/3 .

Probability that Arpit will win = 1/3.

13) Option (c)

Probability of getting an even number of the first dice and the second dice is = 3/6 × 3/6 = 1/4.

14) Option (c)

The maximum total possible = 18

There are two options for Vijay 1) A + B + C = 17

Possibilities = 6, 6, 5

Number of combinations = 3!/2! = 3 2) A + B + C = 18

Possibilities = 6, 6, 6

Number of combinations = 1

Required Probability = 4/216 = 1/52.

15) Option (b)

Probability of all three balls being blue = 5_(C_(3 ) )/9_(C_(3 ) ) = 5/52 Odds in favour of being blue = 5/((42-5) ) = 5 : 37.

16) Option (d) From the diagram, suppose the MD is positioned at say, point 2, then the CFO can be positioned at any of the points 6, 7, 8, 9, 10 Number of favourable positions = 5. Total number of available positions = 11 Thus the required probability = 5/11.

17) Option (d)

7 can be drawn in the following ways? 1 + 6, 2 + 5, 3 + 4, 6 + 1, 5 + 2 and 4 + 3.

Total number of ways = 6

Total number of possibilities = 6 × 6

Probability of not drawing a seven = ((36-6))/36 = 30/36

Odds against drawing a 7 is = 30 : 6 or 5 : 1.

18) The power cycle of 7 is 7,9,3,1

Thus, the only sums possible of 7p+7q is 2,4,6,0,8.

Only when the units digit is 0, is the expression divisible by 5. Hence, the required probability is 1/5.

19) Option (c)

Out of the 12 nuts, 4 are defective and out of the 24 bolts, 8 are defective the probability of picking up both non-defective tools

= 24/36 × 23/135 = 49/105.

20) Option (c)

If the first square chosen is one of the 4 corner squares, the second square can be chosen in 1 way = 4×1.

If the first square chosen is one of the squares on the sides (other than corners) = 24,

the second square can be chosen in 2 ways = 24 × 2

If the first square is any of the middle squares = 36,

the second square can be chosen in 4 ways = 36 × 4.

Total number of ways = 4 + 48 + 144 = 196

Number of ways in which 2 random squares can be selected in a chess board = 64 × 63

Required probability = 196/((64 ×63) ) = 0.048.

21) Option (b)

Total number of ways in which centres can be allocated = 74. Two centres can be chosen in 7C2 ways. The number of ways they can accommodate 4 students = 24 But this includes the two cases where one centre is having all the four students. So, no. of ways = 24 – 2 = 14. So, total number of favourable ways = 7C2 × 14 So, probability = 7C2 × 14/74 = 6/49.

22) Ooption(b)

The number of ways that no two girls are placed together is the number of ways in which 3 places marked with G are selected out of the SEVEN places. This can be done in 7C3ways. Total no. of ways in which balls can be arranged = 10!/7! ×3!= 10C3. So, required probability = 7C3/10C3=7/24.

23) Option (b)

An integer can end with any of the ten’s digits (0, 1, 2 … 9) out of which if it ends with one of the four (0, 1, 5, 6), the required condition will be satisfied.

The probability of an integer ending with 0 or 1 or 5 or 6 is 4/10=2/5.

Now the probability of second integer also ending with the digit that has come in the unit’s place of the first integer is 1/10.

Therefore, the required probability = (2/5)×(1/10)=1/25.

24) Option (c)

For it to be an actual six two things are required to happen together.

1. He is speaking the truth. Let this be an event A.

2. The die shows 6. Let this be the event B.

We have: P(A) = 3/4 P(B) = 1/6

The probability that both events happen together = P(A) × P(B) = 3/4 × 1/6 = 1/8.

25) Option (d)

An experiment succeeds twice as often as it fails. So, the probability of success is 2/3 and the probability of failure is1/3.

In the next 5 trials, the experiment needs to succeed in 4 out of the 5 trials. 4 out of 5 trials can be selected in 5C4 = 5 ways.

So, required probability = 5 × (2/3)4× (1/3) = 80/243.

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