Certainly! Let's solve this question step-by-step.
### Problem Statement:
We are given that the probability of an ape dying when given a newly discovered drug is [tex]\(\frac{1}{6}\)[/tex]. We have selected 6 apes for a trial.
We need to determine the probabilities for the following scenarios:
(a) Half of the apes survived.
(b) More than two-thirds of the apes survived.
(c) Less than one-third of the apes died.
Let's denote:
- The probability that an ape dies as [tex]\( p = \frac{1}{6} \)[/tex].
- The probability that an ape survives as [tex]\( 1 - p = \frac{5}{6} \)[/tex].
- The number of apes [tex]\( n = 6 \)[/tex].
### (a) Probability that half of them survived
To find the probability that half of the apes survived, we need to consider that 3 apes survived (and hence 3 apes died).
By using the binomial distribution:
[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
where [tex]\( n = 6 \)[/tex], [tex]\( k = 3 \)[/tex], [tex]\( p = \frac{1}{6} \)[/tex], and [tex]\( 1-p = \frac{5}{6} \)[/tex]
Therefore, the probability that exactly 3 out of 6 apes survived is:
[tex]\[ P(\text{3 out of 6 survived}) \approx 0.0536 \][/tex]
### (b) Probability that more than two-thirds survived
Two-thirds of 6 is 4. Thus, we need to find the probability that more than 4 apes survived. This includes the cases where 5 or 6 apes survived.
We sum the probabilities for 5 and 6 out of 6 apes surviving:
[tex]\[ P(X > 4) = P(X = 5) + P(X = 6) \][/tex]
By using the binomial distribution:
[tex]\[ P(\text{more than 4 out of 6 survived}) \approx 0.7368 \][/tex]
### (c) Probability that less than one-third died
One-third of 6 is 2. We look for the probability that less than 2 apes died, which means 0 or 1 apes died.
Sum the probabilities for 0 and 1 out of 6 apes dying:
[tex]\[ P(X < 2) = P(X = 0) + P(X = 1) \][/tex]
By using the binomial distribution:
[tex]\[ P(\text{less than 2 out of 6 died}) \approx 0.7368 \][/tex]
### Summary
The probabilities are:
(a) Probability that half of them survived: 0.0536
(b) Probability that more than two-thirds survived: 0.7368
(c) Probability that less than one-third died: 0.7368
These results reflect the specific nature of the probabilities given the conditions.
