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2. Question:In a deck of 52 playing cards, three cards are drawn without replacement. Find the probability that the firs
card is a King, the second card is a Queen, and the third card is a Jack.
Answer:To solve this problem, we'll break down the probability calculation step by step.
1. Probability of drawing a King as the first card: There are 4 Kings in a deck of 52 cards, so the probability of drawi
a King as the first card is 4/52.
2. Probability of drawing a Queen as the second card: After drawing the King, there are 51 cards left in the deck,
including 4 Queens. Thus, the probability of drawing a Queen as the second card is 4/51.
3. Probability of drawing a Jack as the third card: After drawing the King and Queen, there are 50 cards left, includin
4 Jacks. So, the probability of drawing a Jack as the third card is 4/50.
Since these events are independent (drawing one card doesn't affect the probability of drawing another), we can
multiply the probabilities:
Total Probability = (4/52) * (4/51) * (4/50).
Simplify the fractions and calculate the result. The final probability is the product of these three fractions.

3. Question 1: A box contains 10 red balls and 15 blue balls. Three balls are drawn without replacement.
Calculate the probability of drawing 2 red balls followed by 1 blue ball.
Answer 1: To find the probability of this sequence, we break it down into steps. The probability of drawing
the first red ball is 10/25. After that, the probability of drawing a second red ball, without replacement, is
9/24. Lastly, the probability of drawing a blue ball, out of the remaining 23 balls, is 15/23. Since these events
are independent, we multiply these probabilities: (10/25) * (9/24) * (15/23) = 1350 / 13800 = 25 / 256 ≈
0.0977.
Question 2: A fair six-sided die is rolled four times. Find the probability of getting exactly two 3s and two 5s.
Answer 2: The probability of rolling a 3 on a fair six-sided die is 1/6, and the probability of rolling a 5 is also
1/6. The probability of getting exactly two 3s and two 5s in four rolls can be calculated using the binomial
distribution formula: P(X = k) = (n choose k) * p^k * q^(n-k), where n is the number of trials, k is the number
of successful outcomes, p is the probability of success, and q is the probability of failure. In this case, n = 4,
k = 2, p = 1/6, and q = 5/6. Plugging these values into the formula gives P(X = 2) = (4 choose 2) * (1/6)^2 *
(5/6)^2 = 6 * 1/36 * 25/36 = 150 / 1296 ≈ 0.1157.

4. Question 3: An urn contains 8 red balls and 12 green balls. If 5 balls are drawn without replacement, what is the
probability of drawing at least 2 red balls?
Answer 3: To find the probability of drawing at least 2 red balls, we consider two cases: drawing exactly 2 red balls and
drawing 3 red balls. The probability of drawing exactly 2 red balls can be calculated using combinations: (8 choose 2) *
(12 choose 3) / (20 choose 5). Similarly, the probability of drawing 3 red balls is (8 choose 3) * (12 choose 2) / (20
choose 5). Add these two probabilities to get the total probability of drawing at least 2 red balls.
Question 4: A spinner is divided into 8 equal sectors, numbered 1 through 8. If the spinner is spun twice, what is the
probability that the sum of the two numbers is odd?
Answer 4: There are 4 odd and 4 even numbers on the spinner. To find the probability of the sum of two numbers being
odd, we consider two cases: odd + odd and even + even. For odd + odd, the probability is (4/8) * (3/8) = 12/64. For even
+ even, the probability is (4/8) * (3/8) = 12/64. Summing these probabilities gives 24/64 = 3/8.
Question 5: A bag contains 5 red marbles, 4 green marbles, and 6 blue marbles. If 3 marbles are drawn without
replacement, what is the probability of drawing 2 red marbles and 1 green marble?
Answer 5: The probability of drawing 2 red marbles and 1 green marble can be calculated using combinations. The total number
of ways to draw 3 marbles out of 15 is (15 choose 3). The number of ways to draw 2 red marbles out of 5 is (5 choose 2), and the
number of ways to draw 1 green marble out of 4 is (4 choose 1). Multiply these combinations and divide by the total
combinations to get the probability. The result will be (5 choose 2) * (4 choose 1) / (15 choose 3).

5. Question 6: In a deck of 40 cards, there are 10 red cards, 15 blue cards, and 15 green cards. If 4 cards are drawn
without replacement, what is the probability of drawing 2 red cards and 2 blue cards?
Answer 6: To calculate the probability of drawing 2 red cards and 2 blue cards, we need to use combinations.
The total number of ways to draw 4 cards out of 40 is (40 choose 4). The number of ways to draw 2 red cards
out of 10 is (10 choose 2), and the number of ways to draw 2 blue cards out of 15 is (15 choose 2). Multiply
these combinations and divide by the total combinations: [(10 choose 2) * (15 choose 2)] / (40 choose 4).
Simplify to find the probability.
Question 7: A box contains 6 red balls and 8 black balls. Three balls are drawn with replacement. What is the
probability of getting exactly 2 red balls?
Answer 8: Since the balls are drawn with replacement, each draw is independent and doesn't affect the
probabilities of subsequent draws. The probability of drawing a red ball is 6/14, and the probability of drawing a
black ball is 8/14. To get exactly 2 red balls in 3 draws, we consider the binomial distribution formula: P(X = k) =
(n choose k) * p^k * q^(n-k), where n is the number of trials, k is the number of successful outcomes, p is the
probability of success, and q is the probability of failure. In this case, n = 3, k = 2, p = 6/14, and q = 8/14. Plug
these values into the formula to calculate the probability.

6. Question 3: A company has 4 software engineers and 6 data analysts. If 3 employees are selected
randomly to form a team, what is the probability that the team consists of 2 software engineers and 1
data analyst?
Answer 3: To find the probability of forming a team with 2 software engineers and 1 data analyst, we
use combinations. The total number of ways to select 3 employees out of 10 is (10 choose 3). The
number of ways to select 2 software engineers out of 4 is (4 choose 2), and the number of ways to
select 1 data analyst out of 6 is (6 choose 1). Multiply these combinations and divide by the total
combinations to get the desired probability.
Question 4: A fair coin is tossed 5 times. What is the probability of getting at least 3 heads?
Answer 4: To calculate the probability of getting at least 3 heads in 5 coin tosses, we consider the
different combinations of successful outcomes: 3 heads, 4 heads, and 5 heads. The probability of
getting a head in a fair coin toss is 0.5, and the probability of getting a tail is also 0.5. Using the
binomial distribution formula, we calculate the probabilities of these individual cases and then add
them up to get the probability of getting at least 3 heads.

7. Thank you