PROBABILITY PRACTICE QUESTIONS : 1-10

Question 1.) If three unbiased coins are tossed simultaneously, their outputs are observed. Then find the probability of getting exactly two heads?

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

(B) `\frac{1}{4}`

(D) `\frac{5}{8}`

Answer. (C)

Solution.

It is a case of tossing unbiased coins, so we will use a formula to find the total number of events.

If three unbiased coins are tossed simultaneously, then the total number of outcomes = `2^{3}` = 8

Hence, the total number of events = 8

Now, the outputs of the toss will be either heads (H) or tails (T), and here, the output events will be,

(HHH), (HTT), (THT), (HHT), (HTH), (THH), (TTH), (TTT)

But here, we need exactly two heads in the outputs. So our favorable events will be,

(HHT), (HTH), (THH)

So, exactly two heads appear in 3 results. Hence, the number of favorable events = 3

Hence, the required probability = Number of favorable events / Total number of events = `\frac{3}{8}`

Question 2.) If two unbiased dice are rolled together and their outputs are observed, then find the probability that the outputs of both dice are even numbers.

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

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

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

Answer. (C)

Solution.

If two unbiased dices are rolled together then the total number of outcomes (events) = 6 × 6 = 36

Now the even numbers on a dice are (2, 4, 6)

The pair of these numbers will be (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6).

So number of favorable events = 9

Alternative method,

Since there are 3 even numbers and when they will make a pair then the number of combinations made by them will be = 3 × 3 = 9

Hence the required probability = `\frac{9}{36}` = `\frac{1}{4}`

Question 3.) There are 8 green 5 red balls in a bag. If two balls are drawn from the bag one by one without replacement then find the probability that first is red and second is green.

(A) `\frac{10}{39}`

(B) `\frac{8}{13}`

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

Answer. (A)

Solution.

Number of red balls in the bag = 5

Number of green balls in the bag = 8

Total number of balls in the bag = 8 + 5 = 13

It is given that the first ball drawn should be red.

So the probability of drawing a red ball from the bag = `\frac{5}{13}`

Now the first ball is not placed back in the bag after it is drawn.

So now the total number of balls in the bag = 13 – 1 = 12

And it is given that the second ball drawn should be green.

So the probability of drawing a green ball from the bag = `\frac{8}{12}` = `\frac{2}{3}`

Hence the required probability = `\frac{5}{13}` × `\frac{2}{3}` = `\frac{10}{39}`

Question 4.) Three students A, B, and C appear in an examination. If their probabilities of passing the examination are `\frac{2}{5}`, `\frac{1}{2}` and `\frac{1}{3}` respectively then find the probability that at most two of them pass the examination.

(A) `\frac{7}{15}`

(B) `\frac{8}{15}`

(D) `\frac{14}{15}`

Answer. (D)

Solution.

Here the word ‘at most’ has been used and it means that it is the maximum limit for the required probability. The condition of at most two means maximum two students can pass the examination. So we will consider the cases of only two students passing the examination, only one student passing the examination and none of them passing the examination. All these cases will be independent to each other.

Let P(A), P(B), and P(C) are the probabilities of A, B, and C for passing the examination and P̅(A), P̅(B), and P̅(C) are the probabilities of A, B, and C for not passing the examination.

Now it is given that,

P(A) = `\frac{2}{5}`

P(B) = `\frac{1}{2}`

P(C) = `\frac{1}{3}`

Now we will evaluate the values of P̅(A), P̅(B), and P̅(C).

P̅(A) = 1 – P(A) = 1 – `\frac{2}{5}` = `\frac{3}{5}`

P̅(B) = 1 – P(B) = 1 – `\frac{1}{2}` = `\frac{1}{2}`

P̅(C) = 1 – P(C) = 1 – `\frac{1}{3}` = `\frac{2}{3}`

1^st case, only two students pass the examination. It means that if A and B are pass then C is fail or if A and C are pass then B is fail or if B and C are pass then A is fail. So the probability for this case is given by,

P(A) × P(B) × P̅(C) + P(A) × P̅(B) × P(C) + P̅(A) × P(B) × P(C)

= `\frac{2}{5}` × `\frac{1}{2}` × `\frac{2}{3}` + `\frac{2}{5}` × `\frac{1}{2}` × `\frac{1}{3}` + `\frac{3}{5}` × `\frac{1}{2}` × `\frac{1}{3}` = `\frac{4}{30}` + `\frac{2}{30}` + `\frac{3}{30}` = `\frac{9}{30}`

2^nd case, only one student passes the examination. It means that if A is pass then B and C are fail or if B is pass then A and C are fail or if C is pass then A and B are fail. So the probability for this case is given by,

P(A) × P̅(B) × P̅(C) + P̅(A) × P(B) × P̅(C) + P̅(A) × P̅(B) × P(C)

= `\frac{2}{5}` × `\frac{1}{2}` × `\frac{2}{3}` + `\frac{3}{5}` × `\frac{1}{2}` × `\frac{2}{3}` + `\frac{3}{5}` × `\frac{1}{2}` × `\frac{1}{3}` = `\frac{4}{30}` + `\frac{6}{30}` + `\frac{3}{30}` = `\frac{13}{30}`

3^rd case, none of the students pass the examination. It means that all A, B and C are fail. So the probability for this case is given by,

P̅(A) × P̅(B) × P̅(C) = `\frac{3}{5}` × `\frac{1}{2}` × `\frac{2}{3}` = `\frac{6}{30}`

Now, the required probability will be the sum of probabilities of all cases.

Hence the required probability = `\frac{9}{30}` + `\frac{13}{30}` + `\frac{6}{30}` = `\frac{28}{30}` = `\frac{14}{15}`

Alternative method,

Here, only three students are appearing in the examination and the condition is of at most two. So we can use the following conditional method also to calculate the required probability,

Required probability = 1 – probability that all pass the examination

So probability that all pass the examination = P(A) × P(B) × P(C) = `\frac{2}{5}` × `\frac{1}{2}` × `\frac{1}{3}` = `\frac{1}{15}`

Required probability = 1 – `\frac{1}{15}` = `\frac{14}{15}`

Question 5.) If four unbiased coins are tossed simultaneously and there outputs are observed. Then find the probability of getting at least two tails in output.

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

(B) `\frac{11}{16}`

(D) `\frac{9}{16}`

Answer. (B)

Solution.

If four unbiased coins are tossed simultaneously then total number of events = `2^{4}` = 16

Now here we need at least two tails in the outputs. It means all those outcomes that have two or more than two tails will be the favorable events and rest of the events that have either one tail or no tail in outcome will be non-favorable events.

So here we will use the formula,

Total number of events = Number of favorable events + Number of non-favorable events

Now for getting the number of non-favorable events we will consider the outcomes that have either no tail or only one tail and such events will be,

(HHHH), (HHHT), (THHH), (HHTH), (HTHH)

So the number of non-favorable events = 5

And the number of favorable events = 16 – 5 = 11

Hence the required probability = `\frac{11}{16}`

Question 6.) One card is drawn at random from a pack of 52 playing cards. What is the probability that the card drawn is not a face card (A face card is the card having the face of a jack or a queen or a king)?

(A) `\frac{10}{13}`

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

(D) `\frac{1}{13}`

Answer. (A)

Solution.

Let P be the probability of not drawing a face card and P̅ be the probability of drawing a face card. Then,

P = 1 – P̅

Since, there are 12 face playing cards among the total 52 playing cards.

Hence P̅ = `\frac{12}{52}` = `\frac{3}{13}`

And the required probability, P = 1 – P̅ = 1 – `\frac{3}{13}` = `\frac{10}{13}`

Question 7.) Two boys A and B appeared in an interview for selection. If the probability of selection of A is `\frac{1}{6}` and that of B is `\frac{2}{5}` then find the probability that both of them are selected.

(A) `\frac{7}{30}`

(B) `\frac{1}{15}`

(D) `\frac{4}{15}`

Answer. (B)

Solution.

Let P(A) be the probability of selection of A and P(B) is the probability of selection of B.

Then as per the information given in the question,

P(A) = `\frac{1}{6}` and P(B) = `\frac{2}{5}`

Since the events of selection of both the candidates are independent.

Hence the required probability = P(A) × P(B) = `\frac{1}{6}` × `\frac{2}{5}` = `\frac{1}{15}`

Question 8.) A problem is given to three students whose chances of solving it are `\frac{1}{2}`, `\frac{3}{8}` and `\frac{5}{9}` respectively. What is the probability that the problem will be solved?

(A) `\frac{11}{18}`

(B) `\frac{7}{24}`

(D) `\frac{5}{8}`

Answer. (C)

Solution.

Here we need only the result regardless of whoever is giving us the result.

So for such a condition, we use a formula,

Probability that the problem will be solved = 1 – Probability that none solves the problem

Probability that first student solves the problem = `\frac{1}{2}`

Probability that first student does not solve the problem = 1 – `\frac{1}{2}` = `\frac{1}{2}`

Probability that the second student solves the problem = `\frac{3}{8}`

Probability that second student not solves the problem = 1 – `\frac{3}{8}` = `\frac{5}{8}`

Probability that third student solves the problem = `\frac{5}{9}`

Probability that third student not solves the problem = 1 – `\frac{5}{9}` = `\frac{4}{9}`

Probability that none of them solves the problem = `\frac{1}{2}` × `\frac{5}{8}` × `\frac{4}{9}` = `\frac{5}{36}`

Probability that the problem gets solved = 1 – `\frac{5}{36}` = `\frac{31}{36}`

Question 9.) In a class, there are 10 boys and 12 girls. If three students are selected at random then find the probability that there is 1 boy and 2 girls in the selected students.

(A) `\frac{3}{22}`

(B) `\frac{3}{11}`

(D) `\frac{3}{7}`

Answer. (D)

Solution.

Total number of students = 10 + 12 = 22

The number of total events will be the number of ways of selecting 3 students out of 22 = ²²C₃ = 1540

Now, as per the given condition, there will be 1 boy and two girls in the 3 selected students.

So the probability of selecting 1 boy out of 10 = ¹⁰C₁ = 10

And the probability of selecting 2 girls out of 12 = ¹²C₂ = 66

So the number of favorable events = 10 × 66 = 660

Hence the required probability = `\frac{660}{1540}` = `\frac{3}{7}`

Question 10.) A card is drawn from a pack of 52 playing cards. Then find the probability of getting either an ace of heart or a king of spade.

(A) `\frac{1}{26}`

(B) `\frac{1}{4}`

(D) `\frac{1}{52}`

Answer. (A)

Solution.

Since the probability of drawing one card from a pack of 52 playing cards = `\frac{1}{52}`

Now the card drawn may be any card. So as per the condition given in the question there will be two cases but only one will satisfy at a time.

Hence the required probability = `\frac{1}{52}` + `\frac{1}{52}` = `\frac{1}{26}`

CHECK THE COMPLETE LIST OF PRACTICE QUESTIONS