There are four children of diff�erent ages in a family. Their mom purchases six di�erent gifts and decides to
divide the gifts randomly between the children. Assuming that a child can receive no gift or multiple gifts,
answer the following questions:
(a) What is the probability that exactly one of the children receives all six gifts?
(b) What is the probability that each child receives at least one gift?
(c) Thinking only about the number of gifts received by each child, how many diff�erent
possibilities are there? [for example, (3; 1; 1; 1) is one possibility where child 1 receives 3 gifts and each
of the other children receive one; or (1; 3; 1; 1) is another possibility where the second child receives
three gifts and the remaining children receive one; yet (0; 0; 6; 0) is another choice where the third child
receives all the gifts. In this example, I have given examples of three possibilities. The problem is
asking for the total number of such possibilities.]
To answer these questions, let's analyze the problem step by step:
(a) What is the probability that exactly one of the children receives all six gifts?
To calculate this probability, we need to consider that there are four different children and six different gifts. The total number of ways to distribute the gifts is 4^6 because each gift has four possible recipients.
Now, out of these total possibilities, let's count how many ways exactly one child receives all six gifts. There are four options for choosing the child who receives all the gifts, and for each of those, there is only one way to distribute the gifts.
So, the probability that exactly one child receives all six gifts is 4/4^6 = 1/4^5 = 1/1024.
(b) What is the probability that each child receives at least one gift?
To calculate this probability, we need to consider that there are four different children and six different gifts. Similar to the previous question, the total number of ways to distribute the gifts is 4^6.
Now, let's count how many ways each child receives at least one gift. In this case, the method of counting becomes a bit more complex. We can use the principle of inclusion-exclusion.
First, we count the total number of ways to distribute the gifts without any restrictions, which is 4^6. Then, we subtract the number of ways in which at least one child does not receive a gift.
Let's count the number of ways in which at least one child does not receive a gift:
- Select one child out of four not to receive any gift: 4C1
- Distribute the remaining six gifts among the three remaining children: 3^6
However, we need to consider that choosing the child(s) who doesn't receive a gift can occur in different ways. So, we multiply the two counts.
Therefore, the number of ways at least one child does not receive a gift is 4C1 * 3^6.
Finally, the probability that each child receives at least one gift is 1 - (4C1 * 3^6) / 4^6.
(c) Thinking only about the number of gifts received by each child, how many different possibilities are there?
To find the total number of different possibilities, we need to consider the number of gifts received by each child independently. Each child can receive between 0 and 6 gifts.
Let's consider the cases for each child separately:
- Child 1: 0 to 6 possibilities
- Child 2: 0 to 6 possibilities
- Child 3: 0 to 6 possibilities
- Child 4: 0 to 6 possibilities
Therefore, the total number of different possibilities is (6+1) * (6+1) * (6+1) * (6+1) = 7^4 = 2401 possibilities.