Probability

To find the probability that all 4 balls in a bag are white given that two balls drawn randomly are white, we proceed with:

1. Defining the Random Variable:
Let be the number of white balls in the bag. Since the bag contains 4 balls, can take values 0, 1, 2, 3, or 4.
Assuming each value is equally likely, so the probability is:

for .

2. Defining the Event:
Let be the event that two balls drawn randomly from the bag are both white.
We need to find , the probability that all 4 balls are white given that the two drawn balls are white.

3. Applying Bayes' Theorem:
By Bayes' Theorem, we have:

4. Calculating :
Since each value of is equally likely:

5. Calculating :
If , all 4 balls in the bag are white. Therefore, if two balls are drawn, both must be white, so:

6. Calculating :
The total probability of drawing two white balls is given by summing over all possible values of :


The probability of drawing two white balls given is the number of ways to choose 2 white balls from white balls divided by the number of ways to choose 2 balls from 4 balls:


Since for (as we cannot choose 2 white balls if there are fewer than 2), the sum simplifies to:


Substituting and , we get:


Now compute the binomial coefficients:


So:

7. Computing the Final Probability:
Now substitute into Bayes' Theorem:

Final Answer:
The probability that all 4 balls are white given that two drawn balls are white is .

Step 1: Find the value of .
The sum of the probabilities must equal 1: Simplify the equation: Solve for : Using the quadratic formula: Hence, .

Step 2: Calculate the mean of .
The mean is calculated as: Substitute the probabilities: Substitute the value of (from the quadratic solution) and calculate the final answer: Thus, the mean of is .

We are provided with the probability distribution for the random variable and need to determine its variance.

Step 1: Determine the value of Since the sum of the probabilities must equal 1, we have: Substituting the given probabilities, we get: Simplify the expression: This can be rearranged as a quadratic in : Using the quadratic formula: with , , and , we have: This gives: Since probabilities cannot be negative, we select .

Step 2: Compute the Expected Value The expected value is given by: Substitute the values: Simplify:

Step 3: Compute The second moment is: Substitute the values: Simplify: Express all terms with a common denominator: so that:

Step 4: Compute the Variance Variance is given by: Substitute the values: Thus, the variance of is:

Step 1: Total Ways to Select Three Numbers
The given set is: The number of ways to choose any three numbers from this set is:

Step 2: Counting Selections that Form an Arithmetic Progression
For three numbers to be in arithmetic progression, they must be of the form: where is the middle term, and is the common difference. The constraints are: - , - . Rewriting: The number of valid sequences is determined by choosing and such that they remain within the set boundaries. Counting possible values of : - The middle term can be any value from 2 to 49. - The common difference can range from 1 to . Summing up all valid pairs:

Step 3: Computing Probability
Simplifying:

Step 4: Conclusion
Thus, the final probability that three randomly selected numbers form an arithmetic progression is:

We need to determine the probability of drawing two cards where: - One card has a composite number.
- One card has a multiple of 3.
Step 1: Identify Composite and Multiple of 3 Cards A standard deck has numbers from 2 to 10 (excluding face cards) in each suit (hearts, diamonds, spades, clubs). - Composite numbers:
- Multiples of 3:
Each number appears in 4 suits, so: - Cards with composite numbers: - Cards with multiples of 3: - Overlap (cards that are both composite and multiples of 3, i.e., 6 and 9): Using set notation: Using the formula for choosing one from each category:

Step 2: Compute Probability Total ways to choose 2 cards from 52:

When rolling two dice, the total number of possible outcomes is: \

Step 1: Understanding Co-prime Numbers
Two numbers are considered co-prime if their greatest common divisor (GCD) is 1. Our task is to count all pairs of numbers (from the dice rolls) where the two values are co-prime.

Step 2: Counting Co-prime Pairs
We list all valid pairs where the numbers are co-prime: If the first die shows 1: Every number is co-prime with 1, so there are 6 favorable outcomes.
If the first die shows 2: The numbers 1, 3, and 5 are co-prime with 2, giving 3 favorable outcomes.
If the first die shows 3: The numbers 1, 2, and 4 are co-prime with 3, giving 4 favorable outcomes.
If the first die shows 4: The numbers 1, 3, and 5 are co-prime with 4, giving 3 favorable outcomes.
If the first die shows 5: The numbers 1, 2, 3, and 4 are co-prime with 5, giving 4 favorable outcomes.
If the first die shows 6: The numbers 1 and 5 are co-prime with 6, giving 2 favorable outcomes. Summing these values:

Step 3: Compute the Probability
The probability is calculated as the ratio of favorable outcomes (co-prime pairs) to total outcomes: Thus, the probability of rolling two numbers that are co-prime is:

We are given two bags: - Bag contains:
- 3 white balls
- 2 red balls
- 5 blue balls
- Bag contains:
- 2 white balls
- 3 red balls
- 5 blue balls
A ball is first chosen randomly from Bag and placed into Bag . Then, a ball is randomly selected from Bag . We need to determine the probability that the selected ball from Bag is red.

Step 1: Define Probabilities for Transfer from Since a ball is chosen at random from Bag :

Step 2: Update Bag Based on Transfer 1. If a white ball is transferred: - Bag now has 3 white, 3 red, 5 blue. - Probability of selecting a red ball from : 2. If a red ball is transferred:
- Bag now has 2 white, 4 red, 5 blue.
- Probability of selecting a red ball from : 3. If a blue ball is transferred: - Bag now has 2 white, 3 red, 6 blue. - Probability of selecting a red ball from :

Step 3: Compute Total Probability Using Law of Total Probability

Final Answer:

Step 1: Analyze the composition of the word "PROBABILITY". The word "PROBABILITY" consists of 11 letters: P, R, O, B, A, B, I, L, I, T, Y. We observe that the letters B and I are repeated. Thus, the distinct letters are: P, R, O, A, B, I, L, T, Y, making 9 distinct letters in total.

Step 2: Calculate the total number of ways to choose 4 letters from 11. The total number of ways to select 4 letters out of the 11 letters (considering repetitions) is: Using the formula for combinations:

Step 3: Calculate the number of ways to select 4 letters with no repetition. To select 4 letters with no repetition, we choose from the 9 distinct letters (P, R, O, A, B, I, L, T, Y): Using the formula for combinations:

Step 4: Apply the complement rule to calculate the probability of at least one repetition. The probability of no repetition is: Now, the probability of at least one repetition is the complement of the probability of no repetition: Simplifying the fraction: Thus, the probability that at least one letter is repeated is .

Step 1: Probability of picking each urn and drawing a red ball.
- Probability of choosing any one urn is .
- From urn : probability of red is .
- From urn : probability of red is .
- From urn : probability of red is .

Step 2: Bayes'Theorem.
Overall probability of drawing a red ball: We want : Hence .

First, calculate the total number of outcomes where the sum is odd. Odd sums can be obtained by the following combinations (first die, second die): - (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), ... - Repeating similarly for other numbers, there are 3 odd combinations for each number on the first die, giving: Next, find the combinations where the sum is both odd and a multiple of 3 (3, 9, 15 possible, but only 3 and 9 can be rolled): - Sums of 3: (1, 2), (2, 1) - Sums of 9: (3, 6), (4, 5), (5, 4), (6, 3) Total favorable outcomes: Thus, the probability that the sum is a multiple of 3 given that it is odd:

Step 1: Identifying Occurrences of 'LI' in Each Word

The given words are:

ABILITY, PROBABILITY, FACILITY, MOBILITY

We count the number of times the pair "LI" appears in each word:

ABILITY contains 'LI' once.

PROBABILITY contains 'LI' twice.

FACILITY contains 'LI' once.

MOBILITY contains 'LI' once.

Thus, the total occurrences of 'LI' in all words:

1 + 2 + 1 + 1 = 5

Step 2: Probability of Selecting Each Word

Since one word is chosen randomly, the probability of choosing any particular word is:

1/4

Step 3: Probability of 'LI' Appearing in the Chosen Word

ABILITY: 1/6 (as it has 6 consecutive letter pairs).

PROBABILITY: 2/10 = 1/5 (as it has 10 consecutive letter pairs).

FACILITY: 1/8 (as it has 8 consecutive letter pairs).

MOBILITY: 1/7 (as it has 7 consecutive letter pairs).

Step 4: Probability of Choosing 'LI' Across All Words

The total probability of 'LI' being selected is:

(1/4 × 1/6) + (1/4 × 2/10) + (1/4 × 1/8) + (1/4 × 1/7)

= 1/24 + 2/40 + 1/32 + 1/28

Converting to a common denominator of 840:

1/24 = 35/840, 2/40 = 42/840, 1/32 = 26.25/840, 1/28 = 30/840

Summing:

(35 + 42 + 26.25 + 30) / 840 = 133.25 / 840

Step 5: Computing Conditional Probability

The required probability is:

P(PROBABILITY | LI) = P(LI in PROBABILITY) / P(LI in any word)

= (1/4 × 2/10) / (133.25/840) = 21/116

Step 6: Conclusion

Thus, the final answer is:

21/116

We are given that a fair coin is tossed 6 times, and we need to find the conditional probability that exactly 3 heads appear given that the first three tosses resulted in at least 2 heads.

Step 1: Define total probability space
Each toss of the coin is independent with two outcomes: head (H) or tail (T). The probability of any specific sequence of 6 tosses occurring is:

Step 2: Compute probability of the given condition
The probability that the first three tosses result in at least 2 heads can be computed by considering the cases: 1. 2. 3. 4. Each of these cases follows a binomial probability distribution: Using the binomial formula:

Step 3: Compute probability of exactly 3 heads given the condition
We need to find , which is given by: where is the event that at least 2 heads occur in the first 3 tosses. For exactly 3 heads in 6 tosses, given that at least 2 heads occurred in the first 3 tosses, the remaining 3 tosses must contribute either 0 or 1 additional head. This follows the binomial probability:

Step 4: Conclusion
Thus, the correct answer is:

Step 1: Define the Probability Distribution
A standard deck of playing cards consists of 52 cards, out of which 4 are aces. The probability of drawing an ace in a single draw is: Since the draws are done with replacement, the probability of drawing an ace remains constant for both draws.

Step 2: Define the Random Variable
The random variable represents the number of aces drawn in two independent trials. Since each draw is independent and follows a Bernoulli process, follows a binomial distribution: where: - (two trials), - (probability of success in a single trial).

Step 3: Compute the Expected Value (Mean)
The mean of a binomial distribution is given by: Substituting values:

Step 4: Conclusion
Thus, the mean of the probability distribution of is:

To solve this problem, we need to understand the probability of two events: and .

First, let's analyze the matrix:

The numbers in the matrix are: 1, 2, 3, 4, 5, 6, 7, 8, and 9. We need to select three elements from this matrix.

Step 1: Calculate Probability

is the event where the sum of the selected three elements is odd.

• Sum of three numbers is odd if 1 or 3 numbers are odd.
• There are 5 odd numbers (1, 3, 5, 7, 9) and 4 even numbers (2, 4, 6, 8).

To find :

• Case 1: All three numbers are odd. Choose 3 out of 5 odd numbers: .
• Case 2: Two even numbers and one odd number. Choose 2 out of 4 even and 1 out of 5 odd numbers: .

Total favorable outcomes for : .

Total ways to choose any 3 numbers from 9: .

.

Step 2: Calculate

is the event of selecting three elements that are in a row or column.

• Rows: 3 options (each row).
• Columns: 3 options (each column).

Total ways to form rows or columns: 3 (rows) + 3 (columns) = 6.

Analyze each row/column:

• Row 1: Sum is 6; Row 2: Sum is 15; Row 3: Sum is 24.
• Column 1: Sum is 12; Column 2: Sum is 15; Column 3: Sum is 18.

Only Row 2 and Column 2 have odd sums.

Total favorable outcomes for : 2 (Row 2 and Column 2).

.

Step 3: Calculate

.

So, the answer is 17/21.