A drug is essential to the profileration of lymphocytes can cause kidney damage in 1% of patients. If the drug is given to 30 patients, find:
A)the probability exactly 1 patient suffers kidnay damage
B)the probabaility that less than 2 patients suffer kidney damage
c) the probability that at least 3 patients suffer kidney damage
d) the expected # of people who will suffer kidney damage
e) the standard deviation of the number of people who will suffer kidney damage
p = .01
binomial distribution, n = 30
A)
p(k) = C(30,k) p^(k) p^(30-k)
for k = 1
p(1) = [30!/(1!29!)]* .01^1 * .99^29
= 30*.01*.747
=.224
B)
That would be zero or one patient. We know p(1) = .224 so find p(0)
p(0) = (30!/0!30!) *.01^0 * .99^30
= 1 * 1 * .740
=.740
so the probability of less than 3 is .224+.740 = .964
C. 1-.964 = 0.036
D. mean = n p = 30*.01 = 0.3
E. variance = n p(1-p) =30*.01*.99 = .297
s = sqrt(.297) = .545
To solve this problem, we can use the binomial probability formula. The formula is given by:
P(x) = C(n, x) * p^x * q^(n-x)
Where:
- P(x) is the probability of exactly x successes
- C(n, x) is the number of combinations of n objects taken x at a time
- p is the probability of success (in this case, the probability of kidney damage)
- q is the probability of failure (1 - p)
- n is the number of trials (in this case, the number of patients)
Let's solve each part step by step:
A) The probability of exactly 1 patient suffering kidney damage:
- n = 30 (number of patients)
- x = 1 (number of patients with kidney damage)
- p = 0.01 (probability of kidney damage)
- q = 1 - p = 1 - 0.01 = 0.99 (probability of no kidney damage)
Using the binomial probability formula, we get:
P(1) = C(30, 1) * 0.01^1 * 0.99^(30-1)
Now, let's calculate it:
P(1) = (30! / (1! * (30-1)!)) * 0.01 * 0.99^29
B) The probability that less than 2 patients suffer kidney damage:
To find this probability, we need to calculate the sum of the probabilities of 0 and 1 patient suffering kidney damage.
P(<2) = P(0) + P(1)
We have already calculated P(1) in part A.
For P(0):
- n = 30 (number of patients)
- x = 0 (number of patients with kidney damage)
- p = 0.01 (probability of kidney damage)
- q = 1 - p = 1 - 0.01 = 0.99 (probability of no kidney damage)
Using the binomial probability formula, we get:
P(0) = C(30, 0) * 0.01^0 * 0.99^(30-0)
Now, let's calculate it:
P(0) = (30! / (0! * (30-0)!)) * 0.01^0 * 0.99^30
Therefore, P(<2) = P(0) + P(1)
C) The probability that at least 3 patients suffer kidney damage:
To find this probability, we need to calculate the sum of the probabilities of 3, 4, 5, ..., 30 patients suffering kidney damage.
P(≥3) = P(3) + P(4) + ... + P(30)
For each value of x from 3 to 30, we use the formula:
P(x) = C(30, x) * 0.01^x * 0.99^(30-x)
Then calculate the sum of these probabilities.
D) The expected number of people who will suffer kidney damage:
The expected number of people who will suffer kidney damage is calculated by multiplying the number of trials by the probability of success:
Expected number = n * p
E) The standard deviation of the number of people who will suffer kidney damage:
The standard deviation is calculated using the formula:
Standard deviation = sqrt(n * p * q)
Now you have the steps to solve each part of the problem. You can use these formulas and calculations to find the answers.