Question

Find the probability of at least three

successes in six trials of a binomial
experiment in which the probability of
success is 50%.
Round to the nearest tenth of a
percent.

Answer 1

To find the probability of at least three successes in six trials of a binomial experiment where the success rate is 50%, we'll need to consider the complement of this event, which is easier to calculate in this situation. The complement consists of the probability of either 0, 1, or 2 successes in the six trials. By finding the sum of these probabilities, we can subtract it from 1 to find the probability of the original event (3 or more successes).

First, let's recall the formula for the binomial distribution:

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

where:
- P(X = k) is the probability of k successes in n trials,
- C(n, k) is the number of combinations of n items taken k at a time, it can be calculated using the formula C(n, k) = n! / (k! * (n - k)!),
- p is the probability of success for each trial,
- (1 - p) is the probability of failure for each trial,
- n is the number of trials, and
- k is the number of successes.

Since the success probability is 50%, or 0.5, and the complement includes the probability of 0, 1, or 2 successes, we can calculate each of these probabilities.

For k = 0 (zero successes):
P(X = 0) = C(6, 0) * (0.5)^0 * (0.5)^(6 - 0)
P(X = 0) = (6! / (0! * 6!)) * 1 * (0.5)^6
P(X = 0) = 1 * (0.5)^6
P(X = 0) = (1/64)

For k = 1 (one success):
P(X = 1) = C(6, 1) * (0.5)^1 * (0.5)^(6 - 1)
P(X = 1) = (6! / (1! * 5!)) * (0.5) * (0.5)^5
P(X = 1) = 6 * (0.5) * (0.5)^5
P(X = 1) = 6 * (1/64)

For k = 2 (two successes):
P(X = 2) = C(6, 2) * (0.5)^2 * (0.5)^(6 - 2)
P(X = 2) = (6! / (2! * 4!)) * (0.5)^2 * (0.5)^4
P(X = 2) = (15) * (0.25) * (0.0625)
P(X = 2) = 15 * (1/64)

Now we sum up these probabilities to get the complement:
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
P(X < 3) = (1/64) + 6*(1/64) + 15*(1/64)
P(X < 3) = (1 + 6 + 15) / 64
P(X < 3) = 22 / 64
P(X < 3) = 11 / 32

Now to find the probability of at least three successes (P(X >= 3)), we subtract the complement from 1:
P(X ≥ 3) = 1 - P(X < 3)
P(X ≥ 3) = 1 - (11 / 32)
P(X ≥ 3) = (32 / 32) - (11 / 32)
P(X ≥ 3) = 21 / 32

Converting this to a percentage and rounding to the nearest tenth of a percent:
P(X ≥ 3) ≈ (21 / 32) * 100
P(X ≥ 3) ≈ 65.625%

Rounded to the nearest tenth of a percent, the probability is 65.6%.

Answer 2

`(21)/(32)=0.65625`

Explanation:

If the probability of success is 50%, then p=0.5 and q=1-0.5=0.5.

At least three successes in six trials of a binomial experiment means that favorable are 3 successes, 4 successes, 5 successes and 6 successes.

1. 3 successes:

`Pr_1=C^3_6p^3q^(6-3)=(6!)/(3!(6-3)!)\cdot (0.5)^3\cdot (0.5)^3=20\cdot (1)/(2^6)=(5)/(16)`

2. 4 successes:

`Pr_2=C^4_6p^4q^(6-4)=(6!)/(4!(6-4)!)\cdot (0.5)^4\cdot (0.5)^2=15\cdot (1)/(2^6)=(15)/(64)`

3. 5 successes:

`Pr_3=C^5_6p^5q^(6-5)=(6!)/(5!(6-5)!)\cdot (0.5)^5\cdot (0.5)^1=6\cdot (1)/(2^6)=(3)/(32)`

4. 6 successes:

`Pr_4=C^6_6p^6q^(6-6)=(6!)/(6!(6-6)!)\cdot (0.5)^6\cdot (0.5)^1=1\cdot (1)/(2^6)=(1)/(64)`

Now, the probability of at least three successes in six trials of a binomial experiment is

`Pr=Pr_1+Pr_2+Pr_3+Pr_4=(5)/(16)+(15)/(64)+(3)/(32)+(1)/(64)=(20+15+6+1)/(64)=(42)/(64)=(21)/(32)=0.65625`