PROBABILITY PRACTICE QUESTIONS : 21-30

Question 21.) A company receives 50 job applications for 20 vacancies. 90% of the applicants are shortlisted for the interview. But only 15 candidates qualify the interview and remain get rejected. So the company rearranges another interview session for the rejected candidates with changed eligibility criteria. Now only 80% of the rejected candidates are shortlisted for interview for rest of the 5 vacancies. But this time only 4 candidates qualify the interview and remain get rejected. Now the company decides to arrange a last interview session with rest of the candidates with some new changes in the eligibility criteria. But now only 50% of the rejected candidates are shortlisted for interview to fill up the last vacancy. Now find the probability for the selection of each candidate in this last session.

(A) `\frac{1}{12}`

(B) `\frac{1}{20}`

(C) `\frac{1}{10}`

(D) `\frac{1}{2}`

Answer. (C)

Solution.

Since it is given that among the 50 applicants company shortlisted only 90% applicants.

So number of shortlisted applicants = 50 × `\frac{90}{100}` = 45

Now only 15 candidates get selected and remain gets rejected.

So number of rejected candidates = 45 – 15 = 30

Now for the second session of the interview 80% of the rejected candidates are shortlisted.

So number of shortlisted candidates for the second session of interview = 30 × `\frac{80}{100}` = 24

Now only 4 candidates get selected and remain gets rejected.

So number of rejected candidates = 24 – 4 = 20

Now for the third session of the interview 50% of the rejected candidates are shortlisted.

So number of shortlisted candidates for the third session of interview = 20 × `\frac{50}{100}` = 10

So total number of events = 10

Hence the probability of selecting 1 candidate among 10 candidate = `\frac{1}{10}`

Question 22.) Four persons independently solve a certain problem correctly with probabilities of `\frac{1}{2}`, `\frac{3}{4}`, `\frac{1}{4}`, `\frac{1}{8}`. Then find the probability that the problem is solved correctly by at least one of them.

(A) `\frac{3}{256}`

(B) `\frac{21}{256}`

(C) `\frac{235}{256}`

(D) `\frac{253}{256}`

Answer. (C)

Solution.

Let P is the probability of at least one of them solves correctly and P̅ is the probability of none of them solves correctly. Then,

P = 1 – P̅

Probability that the first person does not solve it correctly = 1 – `\frac{1}{2}` = `\frac{1}{2}`

Similarly, for other three persons the probabilities that they do not solve the question correctly are,

1 – `\frac{3}{4}` = `\frac{1}{4}`

1 – `\frac{1}{4}` = `\frac{3}{4}`

1 – `\frac{1}{8}` = `\frac{7}{8}`

Hence, the required probability is, 1 – (`\frac{1}{2}` × `\frac{1}{4}` × `\frac{3}{4}` × `\frac{7}{8}`) = 1 – `\frac{21}{256}` = `\frac{235}{256}`

Question 23.) There are three boxes B1, B2 and B3. First box B1 contains 1 white ball, 3 red balls and 2 black balls, the second box B2 contains 2 white balls, 3 red balls and 4black balls and the third box B3 contains 3 white balls, 4 red balls and 5 black balls. If 1 ball is drawn from each of the boxes B1, B2 and B3, the probability that all 3 drawn balls are not of the same colour is.

(A) `\frac{283}{324}`

(B) `\frac{45}{324}`

(C) `\frac{279}{324}`

(D) `\frac{41}{324}`

Answer. (A)

Solution.

Let probability that all the balls drawn are not of same color is P and probability that all balls are of sane color is P̅

P = 1 – P̅

Probability that all balls are white = `\frac{1}{6}` × `\frac{2}{9}` × `\frac{3}{12}` = `\frac{6}{648}`

Probability that all balls are red = `\frac{3}{6}` × `\frac{3}{9}` × `\frac{4}{12}` = `\frac{36}{648}`

Probability that all balls are black = `\frac{2}{6}` × `\frac{4}{9}` × `\frac{5}{12}` = `\frac{40}{648}`

P̅ = `\frac{6}{648}` + `\frac{36}{648}` + `\frac{40}{648}` = `\frac{82}{648}`

Hence required probability, P = 1 – P̅ = 1 – `\frac{82}{648}` = `\frac{566}{648}` = `\frac{283}{324}`

Question 24.) If three boys and two girls stand in a queue then find the probability, that the number of boys ahead of any girl is at least one more than the number of girls ahead of that particular girl.

(A) `\frac{3}{4}`

(B) `\frac{1}{3}`

(C) `\frac{2}{3}`

(D) `\frac{1}{2}`

Answer. (D)

Solution.

Let B stands for boys and G stands for girls. Then the possible events are,

(BBBGG), (BBGBG), (BBGGB), (BGBGB), (BGGBB), (BGBBG), (GBBBG), (GBBGB), (GBGBB), (GGBBB).

These are total 10 events. Now the required condition will be fulfilled only either a girl will start the sequence or will be at second position and will not acquire the last position as well. So our favourable events will be,

(BBGBG), (BBGGB), (BGBBG), (GBBBG), (GBBGB)

These are total five events.

Hence required probability = `\frac{5}{10}` = `\frac{1}{2}`

Question 25.) There are three boxes numbered as box 1, box 2 and box 3. The box 1 contains three balls numbered as 1, 2, 3; box 2 contains five balls numbered as 1, 2, 3, 4, 5; and box 3 contains seven balls numbered as 1, 2, 3, 4, 5, 6, 7. A ball is drawn from each of the boxes. Let B_n be the number on the ball drawn from the n^th box, n = 1, 2, 3. The probability that the sum B₁ + B₂ + B₃ will be an odd number is

(A) `\frac{29}{105}`

(B) `\frac{53}{105}`

(C) `\frac{57}{105}`

(D) `\frac{1}{2}`

Answer. (B)

Solution.

The sum of three numbers will be an odd number only if all of them are odd or only one of them is odd and other two are even. Now let’s consider the 1^st case when all three numbers are odd then B₁ will have 2 events, B₂ will have 3 events and B₃ will have 4 events. So total number of events will be,

f(E₁) = 2 × 3 × 4 = 24

Now let’s consider the 2^nd case when two numbers are even and only one is odd then there will be three different types of events. In first type of event, B₁ will be even while B₂ andB₃ will be odd. Similarly in other two events B₂ will be even while B₁ andB₃ will be odd and B₃ will be even while B₁ and B₂ will be odd. So all favorable events will be,

f(E₂) = 2 × 2 × 3 + 1 × 3 × 3 + 1 × 2 × 4 = 12 + 9 + 8 = 29

Total number of favorable events will be = 24 + 29 = 53

Total number of possible events will be = 3 × 5 × 7 = 105

Hence the required probability will be = `\frac{53}{105}`

Question 26.) A fair die is rolled repeatedly again and again until a six is obtained as the result. Let X denotes the number of rolls required. The probability that X ≥ 3 is

(A) `\frac{25}{36}`

(B) `\frac{125}{216}`

(C) `\frac{25}{216}`

(D) `\frac{5}{36}`

Answer. (A)

Solution.

Let P(X≥3) be the required probability. Then,

P(X≥3) = 1 – P(X≤2)

Where, P(X≤2) is the probability of getting six in either of the first two events,

P(X≤2) = `\frac{1}{6}` + `\frac{5}{6}` × `\frac{1}{6}` = `\frac{11}{36}`

Hence, the required probability is,

P(X≥3) = 1 – P(X≤2) = 1 – `\frac{11}{36}` = `\frac{25}{36}`

Question 27.) A pair of fair dice is thrown independently three times. The probability of getting a sum of exactly 7 twice is

(A) `\frac{25}{72}`

(B) `\frac{5}{72}`

(C) `\frac{25}{36}`

(D) `\frac{5}{36}`

Answer. (B)

Solution.

If the dices are thrown one time then the favourable events will be (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). So there are total 6 favourable events if only one dice is thrown. Since there are two dices so the total number of events will be,

6 × 6 = 36

So the probability of getting a sum of 7 in first throw is = `\frac{6}{36}` = `\frac{1}{6}`

Hence the probability of getting a sum of 7 twice in three throws is = ³C₂ × `\frac{1}{6}` × `\frac{1}{6}` × `\frac{5}{6}` = `\frac{5}{72}`

Question 28.) A non-leap year is selected at random. Then find the probability that it will have 53 Sundays.

(A) `\frac{3}{7}`

(B) `\frac{2}{7}`

(C) `\frac{1}{7}`

(D) `\frac{4}{7}`

Answer. (C)

Solution.

In a non-leap year, total number of days is 365. Out of them, there are 52 weeks and 1 day extra. Thus, a non-leap year always has 52 Sunday. The remaining 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday.

Out of these 7 cases, we have Sunday in one case.
∴ Total number of outcomes = 7
Number of favorable outcomes = 1

Hence, required probability = `\frac{1}{7}`

Question 29.) Two playing cards are drawn from a pack of 52 playing cards. What is the probability that either both are red or both are kings?

(A) `\frac{63}{221}`

(B) `\frac{3}{26}`

(C) `\frac{7}{13}`

(D) `\frac{55}{221}`

Answer. (D)

Solution.

Total number of events of drawing 2 playing cards from a pack of 52 playing cards = ⁵²C₂ = 1326

Number of events of drawing 2 red playing cards = ²⁶C₂ = 325

So, the probability of drawing two red playing cards is = `\frac{325}{1326}`

Now, the number of events of drawing 2 kings is = ⁴C₂ = 6

So, the probability of drawing two kings is = `\frac{6}{1326}`

 

Now, the number of events of drawing 2 red kings is = ²C₂ = 1

So, the probability of drawing two red kings is = `\frac{1}{1326}`

And the required probability is = `\frac{325}{1326}` + `\frac{6}{1326}` – `\frac{1}{1326}` = `\frac{330}{1326}` = `\frac{55}{221}`

Question 30.) If two different numbers are taken from the set {0, 1, 2, 3, …, 19, 20}; then find the probability that their sum as well as absolute difference are both multiple of 3.

(A) `\frac{1}{10}`

(B) `\frac{4}{15}`

(C) `\frac{3}{10}`

(D) `\frac{7}{18}`

Answer. (A)

Solution.

There will be only one set of numbers that will satisfy the condition given in the question and it is,

{0, 3, 6, 9, 12, 15, 18}

So, we have a total of 7 numbers in the set.

Hence, our favorable events will be = ⁷C₂ = 21

Total number of events = ²¹C₂ = 210

Hence, the required probability is = `\frac{21}{210}` = `\frac{1}{10}`