Question

Some nomadic tribes, when faced with a lifethreatening, contagious disease, try to improve their chances of survival by dispersing into smaller groups. Suppose a tribe of twenty-one people, of whom four are carriers of the disease, split into three groups of seven each. What is the probability that at least one group is free of the disease? (Hint: Find the probability of the complement.)

Answer 1

Answer:

The probability is 0.4842

Step-by-step explanation:

The number of ways or combinations in which we can select k elements  from a group of n elements when the order is not important is calculated as:

$nCk=\frac{n!}{k!(n-k)!}$

Then, probability P that at least one group is free of the disease is calculated as:

P = 1 - P'

Where P' is the probability that all the groups that are formed has the disease. This probability is calculated as a division between the number of ways in which all the groups has a disease and the total number of ways to create the 3 groups.

The total number of ways to create the 3 groups is:

(21C7)*(14C7)*(7C7) = 399,072,960

Because, first, we are going to select 7 people from the 21 to form the 1st group, then, we are going to select 7 people from the 14 that are remaining to form the second group and finally, we need to select 7 people from the last 7 people to form the third group.

At the same way, the number of ways in which we can form 3 groups in which all have the disease are defined by the following events:

• Group 1 has 2 people with the disease and group 2 and 3 has 1 person each. It is calculated as:

  (4C2)(17C5)   *   (2C1)(12C6)    *    (1C1)(6C6)     =   68,612,544

    1st Group         2nd Group           3rd Group

Where 4C2 are the number of ways in which we can select 2 people from the 4 with disease and 17C5 are the number of ways in which we can select 5 people from the 17 that doesn't have the disease. The multiplication of both combinations give as the number of ways in which we can form the first group. For the others groups apply the same logic.

• Group 2 has 2 people with the disease and group 1 and 3 has 1 person each. It is calculated as:

  (4C1)(17C6)   *   (3C2)(11C5)    *    (1C1)(6C6)     =   68,612,544

    1st Group         2nd Group           3rd Group

• Group 3 has 2 people with the disease and group 1 and 2 has 1 person each. It is calculated as:

  (4C1)(17C6)   *   (3C1)(11C6)    *    (2C2)(5C5)     =   68,612,544

    1st Group         2nd Group           3rd Group

So, the probability P' is calculated as:

P' = (68,612,544 + 68,612,544 + 68,612,544) / (399,072,960)

P' = 0.5158

Finally,  probability P that at least one group is free of the disease is calculated as:

P = 1 - P' = 1 - 0.5158 = 0.4842