Probability

To find the probability of getting at least 4 heads when tossing six coins simultaneously, we need to calculate the probability of getting 4, 5, or 6 heads and add them together.

Probability of getting 4 heads: The number of ways to choose 4 heads out of 6 coins can be calculated as: C(6, 4) = 6! / (4!(6 - 4)!) = 15

Probability of getting 5 heads: The number of ways to choose 5 heads out of 6 coins can be calculated as: C(6, 5) = 6! / (5!(6 - 5)!) = 6

Probability of getting 6 heads: The number of ways to choose 6 heads out of 6 coins can be calculated as: C(6, 6) = 6! / (6!(6 - 6)!) = 1

Now, let's calculate the probabilities for each case:

To find the probability of getting at least 4 heads, we add these probabilities together:

Probability of getting at least 4 heads = (15 + 6 + 1) / 64 = 22 / 64 = 11 / 32

Hence, the probability of getting at least 4 heads when six coins are tossed simultaneously is 11/32 or approximately 0.34375, which is approximately 34.375%.

We are given that 1% of the products are defective, and a dealer orders 300 items from the company. We are asked to find the probability that the number of defective items is at most one.
Step 1: The number of defective items in the order follows a binomial distribution with parameters and , as 1% of the items are defective. The probability mass function for a binomial distribution is given by: where is the number of defective items, is the total number of items, is the probability of an item being defective, and is the number of defective items.
Step 2: We need to find the probability that at most one defective item is in the order, i.e., .
Step 3: We use the Poisson approximation for the binomial distribution when is large and is small. The Poisson distribution has the form: where is the mean of the distribution. In this case: Thus, the probability of having 0 defective items is: The probability of having 1 defective item is: Step 4: The probability of having at most one defective item is: Thus, the probability that at most one defective item is drawn is .

We are given the function , and we are asked to find the probability that a number drawn at random from the set lies in the interval where the function is strictly decreasing.
Step 1: The function is strictly decreasing where its derivative is negative. We first find the derivative of : Step 2: To find the intervals where the function is strictly decreasing, we solve the inequality . First, solve the equation to find the critical points: Using the quadratic formula: Thus, the critical points are: Step 3: Now, to determine the sign of , we test the intervals formed by the critical points: , , and .
- For , the function is strictly decreasing, as .
Step 4: We are interested in the set . The numbers in this set correspond to the odd integers between 1 and 59. These numbers are: Thus, the total number of elements in the set is 30.
Step 5: The function is strictly decreasing in the interval , which corresponds to the odd numbers . Thus, there are 4 numbers in the set where the function is strictly decreasing.
Step 6: The probability that a randomly selected number from the set lies in the interval where the function is strictly decreasing is the ratio of favorable outcomes to total outcomes: Thus, the correct answer is .

Step 1: Define the events.
Let B_1 be the event that a bag is chosen from the first set (3 white, 5 black).
Let B_2 be the event that a bag is chosen from the second set (6 white, 4 black).
Let A be the event that a black ball is drawn.

Step 2: Calculate the probabilities of choosing a bag from each set.
There are 2 bags in the first set and 4 bags in the second set, for a total of 6 bags.
P(B_1) =
P(B_2) =

Step 3: Calculate the conditional probabilities of drawing a black ball from each set.
P(A|B_1) = (5 black balls out of 8 total in the first set)
P(A|B_2) = (4 black balls out of 10 total in the second set)

Step 4: Calculate the probability of drawing a black ball. Using the law of total probability:
P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2)
P(A) =
P(A) =

Step 5: Calculate the probability that the black ball came from the first set of bags. Using Bayes' Theorem:
P(B_1|A) =
P(B_1|A) =
However, this is not the answer given. Let's recalculate with the provided answer:
P(B_1|A) =
The answer given is . Let's see if we can get that:
P(B_1|A) =
P(B_1|A) =
However, we are given . Let's find the mistake.
P(A) =
P(B_1|A) =
We have a mistake in the given answer. The correct answer is .

We need to calculate the probability that the sum of three randomly chosen numbers from 1 to 20 is divisible by 3.

Step 1: Understanding Residues Modulo 3 The numbers from 1 to 20 have residues (remainders) when divided by 3:
- Numbers with remainder 0 (divisible by 3): {3, 6, 9, 12, 15, 18}
⇒ 6 numbers - Numbers with remainder 1: {1, 4, 7, 10, 13, 16, 19}
⇒ 7 numbers - Numbers with remainder 2: {2, 5, 8, 11, 14, 17, 20} ⇒ 7 numbers

Step 2: Conditions for Sum to be Divisible by 3 For the sum of three numbers to be divisible by 3, we need one of the following combinations: - All three numbers have the same remainder.
- One number from each residue group.

Step 3: Counting Possible Combinations
Case 1: All numbers have the same remainder
- All zero remainders ⇒ - All one remainders
⇒ - All two remainders
⇒
Total combinations for this case:
Case 2: One number from each residue group Choosing one from each category:

Step 4: Total Favorable Outcomes

Step 5: Total Possible Outcomes The total number of ways to choose 3 numbers from 20 is:

Step 6: Probability Calculation Step 7: Final Answer

Step 1: Determine the total number of ways to distribute the balls.
Each of the 7 balls can be placed into any of the 4 boxes.
Total number of ways = 4⁷.

Step 2: Determine the number of ways to select 3 balls for the first box.
We need to choose 3 balls out of 7 to be placed in the first box.
Number of ways to choose 3 balls = .

Step 3: Determine the number of ways to distribute the remaining 4 balls.
The remaining 4 balls can be placed into any of the other 3 boxes.
Number of ways to distribute the remaining 4 balls = 3⁴ = 81.

Step 4: Calculate the number of favorable outcomes.
Favorable outcomes = .

Step 5: Calculate the probability.
Probability = .
We can rewrite this as:
.
Therefore, the probability that the first box contains 3 balls is .

We are given 5 consecutive natural numbers: . We need to find the probability that for two randomly chosen distinct numbers and , the expression is divisible by 5.

Step 1: Understanding the Condition for Divisibility
From the identity: Since 5 consecutive natural numbers cover all residues modulo 5 (i.e., 0, 1, 2, 3, 4), we will compute the values of .

Step 2: Values of
By Fermat’s Little Theorem:

Step 3: Condition for
- If and , then .
- If and , then .
- If and (or vice versa), then .
Step 4: Probability Calculation
- Total number of ways to choose 2 distinct numbers out of 5: - Number of valid pairs that satisfy (when both residues are equal or both are divisible by 5):

Step 5: Probability Calculation

Final Answer: (C)

Step 1: Define the random variable. Let X be the random variable representing the number of kings drawn.
The possible values of X are 0, 1, and 2.

Step 2: Calculate the probabilities for each value of X.
Total number of ways to draw 2 cards from 52 is .

P(X=0): No kings drawn. Number of ways to choose 2 non-king cards from 48 is .
P(X=0) =

P(X=1): One king drawn. Number of ways to choose 1 king from 4 and 1 non-king from 48 is .
P(X=1) =

P(X=2): Two kings drawn. Number of ways to choose 2 kings from 4 is .
P(X=2) =

Step 3: Calculate the mean of the probability distribution.
Mean ( ) =



Therefore, the mean of the probability distribution of the number of kings is .