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Question 1.) If three unbiased coins are tossed simultaneously, their outcomes are observed. Then find the probability of getting exactly two heads.
(A) `\frac{1}{2}`
(B)
(C)
(D)
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 outcomes of the toss will be either heads (H) or tails (T), and here, the outcome events will be,
(HHH), (HTT), (THT), (HHT), (HTH), (THH), (TTH), (TTT)
But here, we need exactly two heads in the outcomes. 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}`
(C) `\frac{1}{4}`
(D) `\frac{1}{3}`
Answer. (C)
Solution.
If two unbiased dice are rolled together, then the total number of outcomes (events) = 6 × 6 = 36
Now the even numbers on a die are (2, 4, 6)
The pairs of these numbers will be (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6).
So, the number of favorable events = 9
Alternative method,
Since there are 3 even numbers, and when they make a pair, then the number of combinations made by them will be = 3 ×
Hence the required probability = `\frac{9}{36}` = `\frac{1}{4}`
Question 3.) There are 8 green and 5 red balls in a bag. If two balls are drawn from the bag one by one without replacement, then find the probability that the first is red and the second is green.
(A) `\frac{10}{39}`
(B) `\frac{8}{13}`
(C) `\frac{12}{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 =
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}`
(C) `\frac{11}{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 a maximum of 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 of each other.
Let P(A), P(B), and P(C) be the probabilities of A, B, and C passing the examination, and P̅(A), P̅(B), and P̅(C) be the probabilities of A, B, and C not passing the examination.
Now it is given that,
P(A) =
P(B) =
P(C) =
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}`
1st case: only two students pass the examination. It means that if A and B pass, then C fails; or if A and C pass, then B fails; or if B and C pass, then A fails. 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}`
2nd case: only one student passes the examination. It means that if A passes, then B and C fail; or if B passes, then A and C fail; or if C passes, then A and B 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}`
3rd 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 the 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 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 their outputs are observed. Then find the probability of getting at least two tails in the output.
(A) `\frac{5}{16}`
(B) `\frac{11}{16}`
(C) `\frac{5}{8}`
(D) `\frac{9}{16}`
Answer. (B)
Solution.
If four unbiased coins are tossed simultaneously, then the total number of events = `2^{4}` = 16
Now we need at least two tails in the outcomes. It means all those outcomes that have two or more than two tails will be the favorable events, and the rest of the events that have either one tail or no tails in the 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, to get 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 =
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 a card having the face of a jack or a queen or a king)?
(A) `\frac{10}{13}`
(B) `\frac{3}{13}`
(C) `\frac{1}{4}`
(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 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}`
(C) `\frac{7}{15}`
(D) `\frac{4}{15}`
Answer. (B)
Solution.
Let P(A) be the probability of selecting A and P(B) be the probability of selecting 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 selecting both 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}`
(C) `\frac{31}{36}`
(D) `\frac{5}{8}`
Answer. (C)
Solution.
Here we need only the result, regardless of who 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 solve the problem
Probability that first student solves the problem =
Probability that first student does not solve the problem = 1 – `\frac{1}{2}` = `\frac{1}{2}`
Probability that the second student solves the problem =
Probability that second student not solves the problem = 1 – `\frac{3}{8}` = `\frac{5}{8}`
Probability that third student solves the problem =
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 among the selected students.
(A) `\frac{3}{22}`
(B) `\frac{3}{11}`
(C) `\frac{8}{11}`
(D) `\frac{3}{7}`
Answer. (D)
Solution.
Total number of students = 10 + 12 = 22
The total number of 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 among the 3 selected students.
So the number of ways to select 1 boy out of 10 = 10C1 = 10
And the number of ways 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. Find the probability of getting either an ace of hearts or a king of spades.
(A) `\frac{1}{26}`
(B) `\frac{1}{4}`
(C) `\frac{1}{2}`
(D) `\frac{1}{52}`
Answer. (A)
Solution.
Since the probability of drawing one card from a pack of 52 playing cards =
Now the card drawn may be any card. So, under the condition given in the question, there are two cases, but only one can occur at a time.
Hence the required probability = `\frac{1}{52}` + `\frac{1}{52}` = `\frac{1}{26}`
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