Question

A shipment of 8 parrots from Brazil includes 2 parrots with a potentially fatal disease. As usual, the U. S. Customs Office at the shipment's point of entry randomly samples 4 parrots and tests them for disease. What is the probability that the first three parrots selected are healthy

Answer 1

To find the probability that the first three parrots selected are healthy, calculate the product of the probability of selecting a healthy parrot at each step without replacement: (6/8) * (5/7) * (2/3), which results in a probability of 5/14.

Step-by-step explanation:

The task is to calculate the probability that the first three sampled parrots are healthy given that there are a total of 8 parrots, with 2 of them carrying a potentially fatal disease. When a parrot is tested, it is not replaced, which creates a scenario of dependent events where the probability changes after each draw.

To calculate this, consider the total healthy parrots (8-2=6). Initially, the probability that the first parrot is healthy is 6/8. If it is indeed healthy, one healthy parrot has been removed, so the probability that the second parrot is healthy becomes 5/7. If that one is also healthy, the probability that the third parrot is healthy is now 4/6, or 2/3. The combined probability is the product of these individual probabilities: (6/8) * (5/7) * (2/3).

Calculation:

1. 
   
3. Probability of first parrot being healthy: 6/8
4. 
   
6. Probability of second parrot being healthy (given the first is healthy): 5/7
7. 
   
9. Probability of third parrot being healthy (given the first two are healthy): 4/6
10. 
    
12. Total combined probability: (6/8) * (5/7) * (2/3)

Multiplying these probabilities together gives us the probability that the first three parrots selected are healthy: (6/8) * (5/7) * (2/3) which simplifies to 5/14.

1

Answer 2

5 / 14

Step-by-step explanation:

Given that:

Number of parrots shipped = 8

Number of defective parrots = 2

Number of Random samples tested = 4

Number of healthy parrots = 8 - 2 = 6

Probability that first 3 parrots are healthy:

P(1st parrot) * p(2nd parrot) + p(3rd parrot)

P = (number of required outcomes / number of total possible)

USing sampling without replacement :

P(1st parrot healthy ) = 6/ 8 = 3/4

P(2nd parrot healthy) = 5 / 7

P(3rd parrot healthy) = 4 / 6 = 2/3

P(first 3 parrots being healthy) :

(3/4) * (5/7) * (2/3) = 30/84 = 5 / 14