(A) `\frac{1}{12}`
(B)
(C)
(D)
Answer. (C)
Solution.
Since it is given that among the 50 applicants, the company shortlisted only 90% of the 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 the 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 be the probability that at least one of them solves it correctly, and P̅ is the probability that none of them solves it correctly. Then,
P = 1 – P̅
Probability that the first person does not solve it correctly = 1 – `\frac{1}{2}` = `\frac{1}{2}`
Similarly, for the 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 the probability that all the balls drawn are not of the same color be P, and the probability that all balls are of the same color be 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 stand for boys and G stand for girls. Then the possible events are,
(BBBGG), (BBGBG), (BBGGB), (BGBGB),
(BGGBB), (BGBBG), (GBBBG), (GBBGB), (GBGBB), (GGBBB).
These are a total of 10 events. Now the required condition will be fulfilled
only if either a girl starts the sequence or will be at second position and
does not acquire the last position as well. So our favourable events will be,
(BBGBG), (BBGGB), (BGBBG), (GBBBG), (GBBGB)
These are a total of 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. Box 1 contains three balls numbered 1, 2, 3; box 2 contains five balls numbered 1, 2, 3, 4, 5; and box 3 contains seven balls numbered 1, 2, 3, 4, 5, 6, 7. A ball is drawn from each of the boxes. Let Bn be the number on the ball drawn from the nth 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 the other two are even. Now let’s consider the 1st 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 2nd case when two numbers are even and
only one is odd; then there will be three different types of events. In the first
type of event, B1 will be even while B2 andB3
will be odd. Similarly, in the 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 denote 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 dice 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 a total 6 favourable events if only one die 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 the 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, the 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 Sundays. The remaining 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, or 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 =
Now, the number of events of drawing 2 kings is = 4C2 =
6
So, the probability of drawing two kings is =
Now, the number of events of drawing 2 red kings is = 2C2
= 1
So, the probability of drawing two red kings is =
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 their absolute difference are both multiples 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}`
CHECK THE COMPLETE LIST OF PRACTICE QUESTIONS
