Question

Find the probability of each event. A six-sided die is rolled seven times. What is the probability that the die will show an even number at most five times?

Answer 1

To find the probability that a six-sided die will show an even number at most five times when rolled seven times, we need to calculate the sum of the binomial probabilities for getting an even number exactly 0, 1, 2, 3, 4, and 5 times with n=7 and p=1/2.

Step-by-step explanation:

The question involves calculating the probability of a certain outcome when rolling a six-sided die multiple times. Specifically, we want to find the probability that the die will show an even number at most five times when rolled seven times. In this scenario, the possible even numbers are 2, 4, and 6, and each of these has a probability of 1/6 (one out of six possible outcomes), totaling a probability of 1/2 for rolling any even number. To solve this problem, we can use the binomial probability formula:

P(X = k) = C(n, k) * p^k * (1-p)^(n-k)

Where:

C(n, k) is the number of combinations of n things taken k at a time (or "n choose k"),

p is the probability of a single successful outcome,

k is the number of successful outcomes we are interested in,

n is the total number of trials.

In this case, to find the probability that the die shows an even number at most five times, we must consider the probabilities of the die showing an even number exactly 0, 1, 2, 3, 4, and 5 times. We add these probabilities together to get the final probability.

Therefore, the answer involves using binomial probabilities for k=0, 1, 2, 3, 4, and 5 with n=7 and p=1/2.

Answer 2

`(15)/(16)`

Step-by-step explanation:

When a six sided die is rolled, the possible outcomes can be:

{1, 2, 3, 4, 5, 6}

Even numbers are {2, 4, 6}

Odd Numbers are {1, 3, 5}

Probability of even numbers:

`\frac{\text{Favorable cases}}{\text{Total cases }} = (3)/(6) = (1)/(2)`

This is binomial distribution.

where probability of even numbers,
`p =(1)/(2)`

Probability of not getting even numbers (Getting odd numbers)
`q =(1)/(2)`

Probability of getting r successes out of n trials:

`P(r) = _nC_r* p^r q^(n-r)`

Probability of getting even numbers at most 5 times out of 7 is given as:

P(0) + P(1) +P(2) + P(3) +P(4) + P(5)

`\Rightarrow _7C_0* (1)/(2)^0 (1)/(2)^(7)+_7C_1* (1)/(2)^1 (1)/(2)^(6)+_7C_2* (1)/(2)^2 (1)/(2)^(5)+_7C_3* (1)/(2)^3 (1)/(2)^(4)+_7C_4* (1)/(2)^4 (1)/(2)^(3)+_7C_5* (1)/(2)^5 (1)/(2)^(2)`

`\Rightarrow ((1)/(2))^7 (_7C_0+_7C_1+_7C_2+_7C_3+_7C_4+_7C_5)\\`

`\Rightarrow ((1)/(2))^7 (1+7+(7 * 6)/(2)+(7 * 6 * 5)/(3* 2)+(7 * 6 * 5)/(3* 2)+(7 * 6)/(2))\\\Rightarrow (120)/(128) \\\Rightarrow (15)/(16)`