Question 1
The number of tosses of a coin, that are needed so that the probability of getting at least one head is 0.875, is
Explanation
P(at least one head in n tosses) = 1 - P(no heads at all) = 1 - (1/2)n.
Setting this equal to 0.875 (= 78): 1 - (1/2)n = 78, so (1/2)n = 18 = (1/2)3.
So n = 3 tosses.
Setting this equal to 0.875 (= 78): 1 - (1/2)n = 78, so (1/2)n = 18 = (1/2)3.
So n = 3 tosses.
Question 3
A number is selected at random from the first 50 natural numbers. What is the probability that it would be either a two-digit prime number or a composite number lying between 5 and 40?
Explanation
Two-digit prime numbers within 1-50: 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 -- a total of 11 numbers.
Composite numbers strictly between 5 and 40 (i.e. 6 to 39): out of these 34 numbers, 9 are prime (7, 11, 13, 17, 19, 23, 29, 31, 37), so the remaining 34 - 9 = 25 are composite.
Since a number cannot be both prime and composite at the same time, these two groups (11 and 25) do not overlap. Total favourable outcomes = 11 + 25 = 36.
Probability = 3650 = 0.72.
Composite numbers strictly between 5 and 40 (i.e. 6 to 39): out of these 34 numbers, 9 are prime (7, 11, 13, 17, 19, 23, 29, 31, 37), so the remaining 34 - 9 = 25 are composite.
Since a number cannot be both prime and composite at the same time, these two groups (11 and 25) do not overlap. Total favourable outcomes = 11 + 25 = 36.
Probability = 3650 = 0.72.
Question 4
Some dice with six faces have numbers written from Four to Nine. Two such dice are thrown simultaneously. Find the probability that the sum of numbers on the two dice would be 14 or less.
Explanation
Each die shows a value from 4 to 9, giving 6 × 6 = 36 equally likely outcomes in total. It is easier to first count the outcomes where the sum EXCEEDS 14 (i.e. sums of 15, 16, 17, 18), then subtract from 1.
Sum = 18: (9,9) -- 1 way. Sum = 17: (8,9),(9,8) -- 2 ways. Sum = 16: (7,9),(8,8),(9,7) -- 3 ways. Sum = 15: (6,9),(7,8),(8,7),(9,6) -- 4 ways.
Total outcomes with sum > 14 = 1+2+3+4 = 10. So P(sum > 14) = 1036 = 518.
P(sum ≤ 14) = 1 - 518 = 1318.
Sum = 18: (9,9) -- 1 way. Sum = 17: (8,9),(9,8) -- 2 ways. Sum = 16: (7,9),(8,8),(9,7) -- 3 ways. Sum = 15: (6,9),(7,8),(8,7),(9,6) -- 4 ways.
Total outcomes with sum > 14 = 1+2+3+4 = 10. So P(sum > 14) = 1036 = 518.
P(sum ≤ 14) = 1 - 518 = 1318.
Question 5
Three components A, B and C are manufactured separately and then assembled into a finished product. While producing the three components, it is found that 5 percent of component A, 4 percent of component B and 1 percent of component C are defective. What is the probability that the assembled product is free from defects?
Explanation
For the assembled product to be defect-free, all three components must independently be non-defective:
P(A ok) = 1 - 0.05 = 0.95, P(B ok) = 1 - 0.04 = 0.96, P(C ok) = 1 - 0.01 = 0.99.
P(all three ok) = 0.95 × 0.96 × 0.99 = 0.912 × 0.99 ≈ 0.9.
P(A ok) = 1 - 0.05 = 0.95, P(B ok) = 1 - 0.04 = 0.96, P(C ok) = 1 - 0.01 = 0.99.
P(all three ok) = 0.95 × 0.96 × 0.99 = 0.912 × 0.99 ≈ 0.9.
Question 6
Two persons are playing a set of matches. The winner of 4 matches is declared as the winner. Any player has 50% chance to win a match. The probability that the game comes to an end at the fourth match is _______.
Explanation
For the series to end exactly at the 4th match, one player must have already won 4 matches by then -- which is only possible if that player wins ALL 4 of the first 4 matches (any single loss would push the series beyond 4 matches).
P(Player A wins all 4) = (1/2)4 = 116. Similarly, P(Player B wins all 4) = 116.
Since these are the only two ways the series can end exactly at match 4, total probability = 116 + 116 = 216 = 18.
P(Player A wins all 4) = (1/2)4 = 116. Similarly, P(Player B wins all 4) = 116.
Since these are the only two ways the series can end exactly at match 4, total probability = 116 + 116 = 216 = 18.
Question 7
Find the probability that a 3-digit number formed using the digits 1, 3, and 5 (without repetition) is divisible by 3.
Explanation
The digit sum 1 + 3 + 5 = 9 is divisible by 3, and this sum stays the same no matter how the digits are arranged.
So every 3-digit number formed from these digits is divisible by 3 -- the probability is 1.
So every 3-digit number formed from these digits is divisible by 3 -- the probability is 1.
Question 8
Ms. Radhika appeared in interviews at three different companies. In the first company there are 5 candidates, in the second company there are 12 candidates, and in the third company there are 15 candidates. What is the probability that Ms. Radhika would be selected?
Explanation
In each company, exactly one candidate is selected at random from the pool, so P(selected in a company) = 1/(number of candidates there).
P(not selected) at each company: 45, 1112, 1415.
P(not selected anywhere) = 45 × 1112 × 1415 = 616900.
P(selected at least once) = 1 - 616900 = 284900 = 71225.
P(not selected) at each company: 45, 1112, 1415.
P(not selected anywhere) = 45 × 1112 × 1415 = 616900.
P(selected at least once) = 1 - 616900 = 284900 = 71225.
Question 10
If in a class, 50% of the students study mathematics and science, and 70% of the students study mathematics, then the probability of a student studying science given that he/she is already studying mathematics is:
Explanation
P(Science | Mathematics) = P(Science ∩ Mathematics)P(Mathematics) = 0.500.70 = 57.
Question 12
If two dice are rolled, then the probability of getting a greater number on the first die than on the second, given that the sum is equal to 7, is:
Explanation
Outcomes with sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) -- 6 outcomes total.
Of these, the first die shows a greater number in: (4,3), (5,2), (6,1) -- 3 outcomes.
Probability = 36 = 12.
Of these, the first die shows a greater number in: (4,3), (5,2), (6,1) -- 3 outcomes.
Probability = 36 = 12.
Question 14
Two dice are rolled. What will be the probability that one die has a multiple of 3 and the other die has a multiple of 2?
Explanation
For a particular die, P(multiple of 3) = 26 = 13, and P(multiple of 2) = 36 = 12.
Since either die could be the one showing the multiple of 3, there are two equally likely arrangements, so:
P = 2 × 13 × 12 = 13.
Since either die could be the one showing the multiple of 3, there are two equally likely arrangements, so:
P = 2 × 13 × 12 = 13.
Question 16
A card is drawn at random from a well-shuffled deck of 52 cards. What is the probability that the card drawn is either a King or a Heart?
Explanation
|King| = 4, |Heart| = 13, |King ∩ Heart| = 1 (the King of Hearts).
P(King ∪ Heart) = 4 + 13 - 152 = 1652 = 413.
P(King ∪ Heart) = 4 + 13 - 152 = 1652 = 413.
Question 17
In a Shooting competition, A hit the target 6 out of 13 shots, and B hit 8 out of 11 shots. If they both try once, what is the probability that the target would be hit at least once?
Explanation
P(A misses) = 713, P(B misses) = 311.
P(neither hits) = 713 × 311 = 21143.
P(at least one hits) = 1 - 21143 = 122143.
P(neither hits) = 713 × 311 = 21143.
P(at least one hits) = 1 - 21143 = 122143.