PROBABILITY PRACTICE QUESTIONS : 11-20

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

(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 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 successively?

(A) `\frac{3}{8}`

(B) `\frac{5}{16}`

(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 and in a consecutive manner in the outputs. 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 dice is rolled twice, and its outputs are observed. Then find the probability that the outcome appears in the second chance of rolling the dice is greater than the outcome appearing in the first chance of rolling.

(A) `\frac{5}{12}`

(B) `\frac{1}{6}`

(D) `\frac{1}{2}`

Answer. (A)

Solution.

When a dice is rolled twice then the total number of total events = 6 × 6 = 36

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 second chance is greater than the output appeared in rolling the dice in first chance. It means there will be 5 cases and those will be as following,

1^st case, when 1 appears as output in rolling the dice in first chance then we will have 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.

2^nd case, when 2 appears as output in rolling the dice in first chance then we will have 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.

3^rd case, when 3 appears as output in rolling the dice in first chance then we will have following sets of outputs of both chances,

(3, 4), (3, 5), (3, 6)

So here we have got a total of 3 events.

4^th case, when 4 appears as output in rolling the dice in first chance then we will have following sets of outputs of both chances,

(4, 5), (4, 6)

So here we have got a total of 2 events.

5^th case, when 5 appears as output in rolling the dice in first chance then we will have 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}`

(D) `\frac{1}{3}`

Answer. (B)

Solution.

Total number of events = Number of ways to choose 5 pens from set of 10 pens = ¹⁰C₅ = 252

Now there are total 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}`

(D) `\frac{5}{16}`

Answer. (D)

Solution.

If three unbiased coins are tossed simultaneously then 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 will get equal number of heads then either they both will get 0 heads or 1 head or 2 heads or 3 heads.

Now the probability of getting 0 heads in outcome of toss = `\frac{1}{8}`

Now the probability of getting 1 head in outcome of toss = `\frac{3}{8}`

Now the probability of getting 2 heads in outcome of toss = `\frac{3}{8}`

Now the probability of getting 3 heads in outcome of toss = `\frac{1}{8}`

Now the probability for both the persons will be 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}`

(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 = `\frac{4}{52}`

Similarly the probability of drawing a king card = `\frac{4}{52}`

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}`

(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% candidates qualify in the examination and the number of females qualifying the exam is twice the number of males qualifying 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}`

(D) `\frac{17}{60}`

Answer. (B)

Solution.

Since the candidate is being chosen from all the candidates appearing in the examination.

So 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 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}`

(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 × 3 = 12

Now we need only those numbers which are divisible by 5 and the rule for any number to be divisible by 5 is that the it’s 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 on the place of 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 card is an odd numbered card.

(A) `\frac{3}{13}`

(B) `\frac{5}{13}`

(D) `\frac{4}{13}`

Answer. (D)

Solution.

As we know that in the pack of playing cards there are 4 types of cards and each type have 9 number cards. The numbers on cards starts with 2 and ends on 10 and this set has four odd numbers, that are (3, 5, 7, 9). So there are total 16 odd numbered cards.

Now total number of events = Number of ways of drawing a card from 52 cards = ⁵²C₁ = 52

And 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}`