Probability

Binomial distribution: .
Answer: .

: Number of sixes, .
Probability of six on one die: , not six: .
Answer: .

Sum of probabilities = 1:
Expected value:
Check options: , but option (d) is . Recalculate:
Recompute with : Assume different , but calculations confirm . Choose closest or assume option typo.
Answer: (assuming option correction).

This is an example of a Poisson distribution, which is used for modeling the number of events occurring in a fixed interval of time or space.
The average number of defects ( ) per 50 sq.ft. is given as 1. So, the mean of the distribution is .
The Poisson probability formula is .
(a) Probability of no defect:
Here, we need to find . Since and : Given , the probability of no defect is 0.3678.
(b) Probability of at least one defect:
"At least one defect" means . The easiest way to calculate this is by using the complement rule: We already calculated in part (a). The probability of at least one defect is 0.6322.

This is a binomial probability problem.
Number of trials, .
Probability of success (drawing a spade), .
Probability of failure (not drawing a spade), .
The binomial probability formula is .
(a) All the five cards are spades:
Here, . (b) Only 3 cards are spades:
Here, .

This is an example of a binomial distribution.
Let be the random variable representing the number of heads.
The number of trials (tosses), .
The probability of success (getting a head) in a single trial, .
The mean, or expected value , of a binomial distribution is given by the formula .
Substituting the values: The mean number of heads in three tosses of a fair coin is 1.5.
Alternatively, we can list the sample space and calculate the mean directly:
The possible outcomes are: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
Number of heads (X) can be 0, 1, 2, or 3.
P(X=0) [TTT] = 1/8
P(X=1) [HTT, THT, TTH] = 3/8
P(X=2) [HHT, HTH, THH] = 3/8
P(X=3) [HHH] = 1/8
The mean is :

Let be the outcome of the first die and be the outcome of the second die. We want to find the expected value of the sum, . By the linearity of expectation, .
First, find the expected value of a single fair die roll. The possible outcomes are \{1, 2, 3, 4, 5, 6\}, each with a probability of .
The expected value is: Since the second die is also fair, its expected value is the same: .
Now, the expected value of the sum is: