Probability

Let A = event that false coin is chosen. P(A) =
Let B = event of getting 4 heads.
Case 1: False coin chosen: P(B|A) = 1 (since it always shows head)
Case 2: True coin chosen: P(B|A ) =
Total probability of B: Using Bayes' Theorem:

Sample space (given): Outcomes where sum = 6: (1,5), (2,4), (3,3), (4,2), (5,1) ⇒ Total = 5 outcomes Favorable outcomes: (2,4), (4,2) → 2 outcomes where 4 appears at least once Conditional Probability =

The probability of choosing points randomly such that all pairwise distances are less than the radius of the circle can be found using combinatorial geometry.
The area that satisfies the condition is calculated, and the total area is normalized to find the probability.
The final result is .

To calculate the probability, we use combinations to determine the total number of ways to choose 3 students, as well as the favorable outcomes for choosing 2 girls and 1 boy.

- The total number of ways to choose 3 students from 18 is given by .
- The number of ways to choose 2 girls from 8 is .
- The number of ways to choose 1 boy from 10 is .

Thus, the probability is given by the formula: Therefore, the probability is .

The formula for the union of two independent events is: Since and are independent, . Substituting the known values: Simplifying this equation:

For each ant, there are 4 choices of adjacent vertices. Given that there are 6 ants in total, the total number of ways they can move is , as each ant can independently choose one of the 4 adjacent vertices.

To find the number of valid ways where no two ants arrive at the same vertex, we can think of distributing the 6 ants to the 6 vertices in such a way that no two ants occupy the same vertex. This is a permutation problem, where we are arranging 6 distinct ants on 6 vertices, which can be done in ways. However, since the problem specifies that no two ants can be on the same vertex, we only count valid configurations where each ant is placed on a different vertex.

This gives us 21 valid configurations, corresponding to the number of ways to assign the 6 ants to 6 distinct vertices without overlap.

Thus, the probability is: This fraction simplifies to .

Therefore, the probability is .

We are asked to find the probability that at least one of the events , , or occur. The probability of at least one of the events occurring is given by the formula: We are given the following information: - - - Now, substitute these values into the formula: We are not given , but since and , it follows that: Thus, the probability becomes: Therefore, the probability that at least one of the events , , or occur is .

Step 1: Identify possible outcomes.
When two dice are tossed, the total number of possible outcomes is .
Step 2: Identify prime sums.
Possible prime sums are .
Step 3: Count favorable outcomes.
Sum = 2 1 outcome
Sum = 3 2 outcomes
Sum = 5 4 outcomes
Sum = 7 6 outcomes
Sum = 11 2 outcomes
Total favorable outcomes = .
Step 4: Calculate probability.