1.
In a class, there are 12 boys and 16 girls. One of them is called out by an enroll number, what is the probability that the one called is a girl?
Correct Answer
D. 4/7
Explanation
The probability of selecting a girl from the class can be calculated by dividing the number of girls by the total number of students. In this case, there are 16 girls and 28 students in total (12 boys + 16 girls). Therefore, the probability of selecting a girl is 16/28 which simplifies to 4/7.
2.
There are 12 boys and 8 girls in a tution centre. If three of them scored first mark, then what is the probability that one of the three is a girl and the other two are boys?
Correct Answer
C. 44/95
Explanation
The probability of selecting a girl out of the total number of girls is 8/20. The probability of selecting two boys out of the total number of boys is (12/20)*(11/19). Therefore, the probability of selecting one girl and two boys is (8/20)*(12/20)*(11/19). Simplifying this expression gives us 44/95.
3.
Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
Correct Answer
B. 3/4
Explanation
When two dice are thrown simultaneously, there are a total of 36 possible outcomes (6 outcomes for each die). To find the probability of getting two numbers whose product is even, we need to consider the possible combinations. The only way to get an odd product is by having both dice show odd numbers. There are 9 possible outcomes for this (1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5). Therefore, the probability of getting an odd product is 9/36 = 1/4. Since the sum of probabilities must be 1, the probability of getting an even product is 1 - 1/4 = 3/4.
4.
From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?
Correct Answer
C. 1/221
Explanation
The probability of drawing a king from a pack of 52 cards is 4/52, since there are 4 kings in a deck. After the first king is drawn, there are only 51 cards left in the deck, and only 3 kings remaining. Therefore, the probability of drawing a second king is 3/51. To find the probability of both cards being kings, we multiply the probabilities together: (4/52) * (3/51) = 1/221.
5.
Two dice are tossed. The probability that the total score is a prime number is:
Correct Answer
B. 5/12
Explanation
When two dice are tossed, the total score can range from 2 to 12. Out of these possible outcomes, there are 15 prime numbers: 2, 3, 5, 7, 11. To find the probability of getting a prime number as the total score, we need to calculate the number of favorable outcomes (prime numbers) and divide it by the total number of possible outcomes. The favorable outcomes are 5 (the 5 prime numbers mentioned above), and the total number of outcomes is 36 (since there are 6 possible outcomes for each dice, resulting in 6x6=36 possible combinations). Therefore, the probability of getting a prime number as the total score is 5/36, which simplifies to 5/12.
6.
Three unbiased coins are tossed. What is the probability of getting at most two heads?
Correct Answer
D. 7/8
Explanation
When three unbiased coins are tossed, there are a total of 8 possible outcomes: HHH, HHT, HTH, THH, TTH, THT, HTT, TTT. Out of these, 7 outcomes have at most two heads (HHH has three heads). Therefore, the probability of getting at most two heads is 7/8.
7.
A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:
Correct Answer
A. 1/26
Explanation
There are 4 queens of clubs and 4 kings of hearts in a deck of 52 cards. Since there are no duplicates, the probability of drawing either a queen of clubs or a king of hearts is 4/52. Simplifying this fraction gives us 1/13, which is the correct answer.
8.
A bag contains 7 blue balls and 5 yellow balls. If two balls are selected at random, what is the probability that none is yellow?
Correct Answer
B. 7/22
Explanation
The probability of selecting a yellow ball on the first draw is 5/12. After removing one yellow ball from the bag, there are now 16 balls left, including 7 blue balls. Therefore, the probability of selecting a blue ball on the second draw is 7/16. To find the probability of both events happening (selecting no yellow balls), we multiply the probabilities together: (5/12) * (7/16) = 35/192. However, the question asks for the probability of none of the balls being yellow, so we subtract this probability from 1: 1 - 35/192 = 157/192. This fraction can be simplified to 7/22.
9.
A bag contains four blue, three green and five red balls.If three balls are drawn at random from the bag, then what is the probability that all are red?
Correct Answer
B. 1/22
Explanation
The probability of drawing a red ball on the first draw is 5/12. After the first red ball is drawn, there are 4 red balls remaining out of 11 total balls. Therefore, the probability of drawing a red ball on the second draw is 4/11. Similarly, after two red balls are drawn, there are 3 red balls remaining out of 10 total balls. Therefore, the probability of drawing a red ball on the third draw is 3/10. To find the probability of all three balls being red, we multiply these probabilities together: (5/12) * (4/11) * (3/10) = 1/22.
10.
A bag contains four blue, three green and five red balls.If four balls are drawn at random, what is the probability that two are blue and two are green?
Correct Answer
A. 2/55
Explanation
The probability of drawing two blue balls and two green balls can be calculated by multiplying the probability of drawing two blue balls (4/12 * 3/11) and the probability of drawing two green balls (3/10 * 2/9). Simplifying this expression gives us (4/66 * 3/55) = 2/55. Therefore, the probability that two balls are blue and two are green is 2/55.
11.
A bag contains four blue, three green and five red balls
If three balls are drawn at random, what is the probability that all the three balls are of different colours?
Correct Answer
C. 3/11
Explanation
To find the probability that all three balls drawn are of different colors, you can calculate it as follows:
1. Calculate the total number of ways to choose 3 balls out of 12 (4 blue, 3 green, and 5 red). This is a combination, and it can be calculated using the formula C(n, k) = n! / (k!(n-k)!), where n is the total number of balls and k is the number of balls you want to choose. In this case, it's C(12, 3).
C(12, 3) = 12! / (3!(12-3)!) = 220 ways
2. Calculate the number of ways to choose 3 balls with different colors. To do this, you can calculate the combinations for each color separately and then multiply them.
- Choose 1 blue out of 4 (C(4, 1) ways).
- Choose 1 green out of 3 (C(3, 1) ways).
- Choose 1 red out of 5 (C(5, 1) ways).
Then multiply these combinations together: C(4, 1) * C(3, 1) * C(5, 1) = 4 * 3 * 5 = 60 ways.
3. Now, you can calculate the probability that all three balls are of different colors by dividing the number of ways to choose different-colored balls by the total number of ways to choose 3 balls:
Probability = (Number of ways to choose different colors) / (Total number of ways to choose 3 balls)
Probability = 60 / 220
Simplify the fraction:
Probability = 3/11
So, the probability that all three balls drawn are of different colors is 3/11.
12.
A bag contains four blue, three green and five red balls.If three balls are drawn at random, what is the probability that three is at least one red ball in the three balls?
Correct Answer
A. 37/44
Explanation
The probability of drawing at least one red ball can be calculated by finding the probability of not drawing any red balls and subtracting it from 1. The probability of not drawing any red balls is equal to the probability of drawing all blue and green balls, which can be calculated as (4/12) * (3/11) * (2/10) = 1/55. Therefore, the probability of drawing at least one red ball is 1 - 1/55 = 54/55. However, the question asks for the probability of drawing at least one red ball out of three balls, so we need to consider all possible combinations of selecting three balls. The total number of ways to select three balls out of twelve is 12C3 = 220. Therefore, the probability of drawing at least one red ball out of three balls is (54/55) * (220/1) = 37/44.
13.
If 3 cards are drawn simultaneously form a pack of well shuffled cards, find the probability of then being all Queen.
Correct Answer
E. None of these
14.
In a shooting test, the probability of hitting the target is 1/2 for A and 2/3 for B and 3/4 for C. Find the probability that the target is hit
Correct Answer
B. 23/24
Explanation
The probability of hitting the target for A is 1/2, for B is 2/3, and for C is 3/4. To find the probability that the target is hit, we can calculate the complement of the probability that the target is missed. The probability that A misses the target is 1 - 1/2 = 1/2. The probability that B misses the target is 1 - 2/3 = 1/3. The probability that C misses the target is 1 - 3/4 = 1/4. Therefore, the probability that the target is missed by all three shooters is (1/2) * (1/3) * (1/4) = 1/24. The probability that the target is hit is 1 - 1/24 = 23/24.
15.
In a shooting test, the probability of hitting the target is 1/2 for A and 2/3 for B and 3/4 for C. Find the probability that all of them hit the target.
Correct Answer
C. 1/4
Explanation
The probability that all of them hit the target can be found by multiplying the individual probabilities of each person hitting the target. So, the probability for A is 1/2, for B is 2/3, and for C is 3/4. Multiplying these probabilities together gives (1/2) * (2/3) * (3/4) = 1/4. Therefore, the probability that all of them hit the target is 1/4.