Question 11.) There are some white balls and 25 black balls in a bag of balls. A ball is drawn at random. If the probability that the ball drawn is white is `\frac{2}{7}`, then find the total number of balls in the bag.
(A) 35
(B) 30
(C) 42
(D) 49
Answer. (A)
Solution.
Let the number of white balls = a
So the total number of balls in the bag = a + 25
Now the probability of drawing one white ball from the bag = `\frac{a}{a+25}` = `\frac{2}{7}`
On solving the above equation, we get a = 10
Hence, the total number of balls in the bag = 10 + 25 = 35
Question 12.) If an unbiased coin is tossed four times, and the outputs are observed. Then find the probability of getting exactly three heads in succession?
(A) `\frac{3}{8}`
(B) `\frac{5}{16}`
(C) `\frac{1}{8}`
(D) `\frac{1}{4}`
Answer. (C)
Solution.
If an unbiased coin is tossed 4 times, then the total number of outcomes (events) = `2^{4}` = 16
Now, here we need exactly three heads in a consecutive manner in the
outcomes. So our favorable events will be,
(THHH), (HHHT)
So the number of favorable events = 2
Hence the required probability = `\frac{2}{16}` = `\frac{1}{8}`
Question 13.) A fair die is rolled twice, and its outcomes are observed. Then find the probability that the outcome on the second chance of rolling the dice is greater than the outcome appearing on the first chance of rolling.
(A) `\frac{5}{12}`
(B) `\frac{1}{6}`
(C) `\frac{5}{6}`
(D) `\frac{1}{2}`
Answer. (A)
Solution.
When a die is rolled twice, the total number of events = 6 ×
Now, as per the condition given in the question, the favorable events
will be the events when the output appeared in rolling the dice in the second
chance is greater than the output appeared in rolling the dice in the first chance.
It means there will be 5 cases, and those will be as follows,
1st case: when 1 appears as the output in rolling the dice in the first chance, then we will have the following sets of outputs of both chances,
(1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
So here we have got a total of 5 events.
2nd case: when 2 appears as the output in rolling the dice on the first chance, then we will have the following sets of outputs of both chances,
(2, 3), (2, 4), (2, 5), (2, 6)
So here we have got a total of 4 events.
3rd case: when 3 appears as the output when rolling the dice in the first chance, then we will have the following sets of outputs of both chances,
(3, 4), (3, 5), (3, 6)
So here we have got a total of 3 events.
4th case: when 4 appears as the output when rolling the dice in the first chance, then we will have the following sets of outputs of both chances,
(4, 5), (4, 6)
So here we have got a total of 2 events.
5th case: when 5 appears as the output in rolling the dice in the first chance, then we will have the following output in both chances,
(5, 6)
So here we have got only 1 event.
Now the number of favorable events = 5 + 4 + 3 + 2 + 1 = 15
Hence the required probability = `\frac{15}{36}` = `\frac{5}{12}`
Question 14.) A set of 10 pens contains 3 defective pieces. If 5 pens are chosen randomly, then find the probability that only one pen among the 5 chosen pens is defective.
(A) `\frac{1}{2}`
(B) `\frac{5}{12}`
(C) `\frac{5}{6}`
(D) `\frac{1}{3}`
Answer. (B)
Solution.
Total number of events = Number of ways to choose 5 pens from a set of 10 pens = 10C5 = 252
Now there are a total of 3 defective pens and 7 non-defective pens. If one
defective pen is selected in the 5 pens, then there must be 4 non-defective pens.
So the number of ways of selecting exactly one defective pen out of 3 = 3C1 = 3
And the number of ways of selecting 4 non-defective pens out of 7 = 7C4 = 35
Hence the required probability = `\frac{3×35}{252}` = `\frac{5}{12}`
Question 15.) Three unbiased coins are tossed by two persons independently. Find the probability that both of them get an equal number of heads in their outcomes.
(A) `\frac{3}{8}`
(B) `\frac{5}{8}`
(C) `\frac{1}{4}`
(D) `\frac{5}{16}`
Answer. (D)
Solution.
If three unbiased coins are tossed simultaneously, then the total number of events = `2^{3}` = 8
The output events will be,
(HHH), (HTT), (THT), (TTH), (HHT),
(HTH), (THH), (TTT)
Now, as per the condition given in the question, if the two persons tossing
the coins get an equal number of heads, then either they both will get 0 heads, 1 head, 2 heads, or 3 heads.
Now the probability of getting 0 heads in outcome of toss =
Now the probability of getting 1 head in outcome of toss =
Now the probability of getting 2 heads in outcome of toss =
Now the probability of getting 3 heads in outcome of toss =
Now the probability for both persons will
be the same for each event.
Hence required probability = `\frac{1}{8}` × `\frac{1}{8}` + `\frac{3}{8}` × `\frac{3}{8}` + `\frac{3}{8}` × `\frac{3}{8}` + `\frac{1}{8}` × `\frac{1}{8}` = `\frac{5}{16}`
Question 16.) If a card is drawn at random from a pack of 52 playing cards, then find the probability that the card is either a queen or a king.
(A) `\frac{8}{13}`
(B) `\frac{1}{13}`
(C) `\frac{2}{13}`
(D) `\frac{4}{13}`
Answer. (C)
Solution.
Since in the pack of 52 playing cards, there are 4 queen cards and 4 king cards.
So the probability of drawing a queen card =
Similarly the probability of drawing a king
card =
But we need to draw only one card, either a king or a queen.
Hence the required probability = `\frac{4}{52}` + `\frac{4}{52}` = `\frac{8}{52}` = `\frac{2}{13}`
Question 17.) If A and B are two events such that P(`\frac{A}{B}`) = `\frac{3}{10}`, P(`\frac{B}{A}`) = `\frac{4}{7}`, and P(A⋂B) = `\frac{2}{9}`, then find the value of P(A̅⋃B̅).
(A) `\frac{10}{27}`
(B) `\frac{5}{54}`
(C) `\frac{10}{54}`
(D) `\frac{49}{54}`
Answer. (B)
Solution.
Since we know that P(A̅⋃B̅) = 1 – P(A⋃B)
And P(A⋃B) = P(A) + P(B) – P(A⋂B) ………eq.(1)
Now we have the values of P(A⋂B), P(A/B), and P(B/A) given in the question.
We need to find the values of P(A) and P(B), and it is given by the
following formulas,
P(A) = {P(A⋂B)}/{P(A/B)} and P(B) = {P(A⋂B)}/{P(B/A)}
Now, putting values and further calculating we
get,
P(A) = `\frac{20}{27}` and P(B) = `\frac{7}{18}`
Now putting all the values in eq.(1) we get,
P(A⋃B) = `\frac{20}{27}` + `\frac{7}{18}` – `\frac{2}{9}` = `\frac{49}{54}`
P(A̅⋃B̅) = 1 – P(A⋃B) = 1 – `\frac{49}{54}` = `\frac{5}{54}`
Question 18.) In an examination, 120 candidates appear, including both males and females. If only 65% of the candidates qualify in the examination, and the number of females qualifying in the exam is twice the number of males qualifying in it. Now a candidate is randomly chosen from all the appearing candidates. Then find the probability that the chosen candidate is a qualified female candidate.
(A) `\frac{11}{20}`
(B) `\frac{13}{30}`
(C) `\frac{2}{3}`
(D) `\frac{17}{60}`
Answer. (B)
Solution.
Since the candidate is being chosen from all the candidates appearing in the examination.
So, the total number of events = Total number of candidates appearing in the
examination = 120
Number of candidates qualified in examination = 120 × `\frac{65}{100}` = 78
Now, as per the condition given in the
question, the number of female candidates is twice the number of male candidates
among all the qualified candidates. So if ‘x’ is the number of qualified male
candidates, then ‘2x’ will be the number of qualified female candidates.
Now we have,
2x + x = 78
On solving the above equation, we will get
that the number of qualified male candidates = 26
And the number of qualified female candidates
= 52
So, the number of favorable events = Number of
qualified female candidates = 52
Hence the required probability = `\frac{52}{120}` = `\frac{13}{30}`
Question 19.) A 2-digit number is formed using the digits of the set of the first four prime numbers without any repetition. Then find the probability that the number is divisible by 5.
(A) `\frac{1}{4}`
(B) `\frac{1}{3}`
(C) `\frac{1}{12}`
(D) `\frac{1}{2}`
Answer. (A)
Solution.
The first four prime numbers are (2, 3, 5, 7)
Total 2-digit numbers can be formed from these 4 prime numbers without
repetition = 4 ×
Now we need only those numbers which are divisible by 5, and the rule for
any number to be divisible by 5 is that its unit digit must be 0 or 5.
Now we don’t have 0 here, but we have 5, so we need to fix the number 5 in
the place of the unit digit. In this case, we will get only 3 numbers.
So the favorable events = 3
Hence the required probability = `\frac{3}{12}` = `\frac{1}{4}`
Question 20.) If a card is drawn at random from a pack of 52 playing cards, then find the probability that the card is an odd-numbered card.
(A) `\frac{3}{13}`
(B) `\frac{5}{13}`
(C) `\frac{9}{13}`
(D) `\frac{4}{13}`
Answer. (D)
Solution.
As we know, in a pack of playing cards there are 4 types of cards, and each type has 9 number cards. The numbers on the cards start with 2 and end with 10, and this set has four odd numbers, which are (3, 5, 7, 9). So there are a total of 16 odd-numbered cards.
Now, the total number of events = Number of ways of
drawing a card from 52 cards = 52C1 = 52
And the number of favourable events = Number of ways of
drawing a card from 16 odd-numbered cards = 16C1 = 16
Hence the required probability = `\frac{16}{52}` = `\frac{4}{13}`
CHECK THE COMPLETE LIST OF PRACTICE QUESTIONS
