Can anybody help me on this problem for my Algebra 2 class, I am having a hard time tyring to solve it step by step.
The probability that an archer hits the target is
p = 0.8, so the probability that he misses the target is q = 0.2. It is known that in this situation the probability that the archer hits the target exactly r times in n attempts is given by the term containing pr in the binomial expansion of (p + q)^n.
Find the probability that the archer hits the target exactly four times in six attempts
To find the probability that the archer hits the target exactly four times in six attempts, we can use the binomial expansion formula.
The binomial expansion formula states that for any integers n and k, the term containing pr in the binomial expansion of (p + q)^n is given by the formula:
C(n, r) * p^r * q^(n-r),
where C(n, r) represents the number of combinations of choosing r items from a set of n items, and can be calculated using the formula:
C(n, r) = n! / (r! * (n-r)!),
where ! denotes the factorial of a number.
In this question, we have n = 6 (6 attempts) and r = 4 (exactly four hits).
Let's calculate the probability step by step:
1. Calculate C(6, 4):
C(6, 4) = 6! / (4! * (6-4)!)
= 6! / (4! * 2!)
= (6 * 5 * 4 * 3 * 2 * 1) / ((4 * 3 * 2 * 1) * (2 * 1))
= (6 * 5) / (2 * 1)
= 15.
2. Calculate p^r:
p^r = 0.8^4 = 0.8 * 0.8 * 0.8 * 0.8 = 0.4096.
3. Calculate q^(n-r):
q^(n-r) = 0.2^(6-4) = 0.2^2 = 0.04.
4. Multiply all the calculated values:
C(6, 4) * p^r * q^(n-r) = 15 * 0.4096 * 0.04
= 15 * 0.016384
= 0.24576.
Therefore, the probability that the archer hits the target exactly four times in six attempts is approximately 0.24576, or 24.576%.
To find the probability that the archer hits the target exactly four times in six attempts, we can use the binomial distribution formula. The formula is:
P(X = k) = C(n, k) * p^k * q^(n-k)
Where:
- P(X = k) is the probability of getting exactly k successes in n trials
- C(n, k) is the number of combinations of n items taken k at a time, which can be calculated using n! / (k! * (n-k)!)
- p is the probability of success on a single trial
- q is the probability of failure on a single trial (1 - p)
- n is the total number of trials
- k is the number of successful trials we want to find the probability for
Now, we can substitute the given values into the formula:
p = 0.8 (probability of hitting the target)
q = 0.2 (probability of missing the target)
n = 6 (total number of attempts)
k = 4 (number of successful attempts we want to find the probability for)
Let's calculate the probability step by step:
First, let's calculate the number of combinations:
C(6, 4) = 6! / (4! * (6-4)!) = 6! / (4! * 2!) = (6 * 5 * 4 * 3) / (4 * 3 * 2 * 1) = 15
Next, let's calculate the probability of hitting the target four times:
p^k = 0.8^4 = 0.4096
Now, let's calculate the probability of missing the target (2 attempts):
q^(n-k) = 0.2^(6-4) = 0.2^2 = 0.04
Finally, let's calculate the overall probability:
P(X = 4) = C(6, 4) * p^k * q^(n-k) = 15 * 0.4096 * 0.04 = 0.24576
Therefore, the probability that the archer hits the target exactly four times in six attempts is 0.24576.