Mrs. Napholtz's Math Site

Probability Problems

In a math class, there are four students in the first row: three boys, Arthur, David and Carlos, and one girl, Kim. The teacher will call one of these students to the board to solve a problem. When the problem is solved, the teacher will then call upon one of the remaining students in the first row to do a second problem at the board.

• Draw a tree diagram or list the sample space (all possible outcomes) for calling two students to the board.
• Find the probability that:
  1. Kim will be one of the two students called to the board.
  2. At least one boy will be called to the board.
  3. Kim and Arthur will be the two students called to the board.
  4. Two girls will be called to the board.

Sample Space:

Probabilities:

1. Kim appears in 6 out of 12 possibilities: P = 1/2
2. At least one boy appears in ALL of the possible outcomes: P = 1
3. Kim and Arthur appear together in 2 of the 12: P = 1/6
4. There can never be two girls: P = 0

This problem uses probability and combinations. You may use the following formula or your calculator to compute _nC_r:
_nC_r = n! / [r! (n - r)! ]
To compute _nC_r using the TI-82 or TI-83s, enter the "n" number, hit MATH, then cursor to the PRB menu. From that menu, choose _nC_r then enter the "r" number and hit enter.
Here's the problem:

A committee of 6 is to be chosen from 4 juniors and 5 seniors.

1. What is the probability that the committee will include the same number of juniors and seniors?
2. What is the probability that the committee will include all 5 seniors?
3. What is the probability that the committee will include no seniors?

HINTS:

• The total number of possible outcomes is ₉C₆ Note that Answer to # 1 = [(ways to choose 3 jrs)(ways to choose 3 srs)]/total outcomes
• Question 3 is very easy.

1. P(3 juniors, 3seniors) = [(ways to choose 3 jrs)(ways to choose 3 srs)]/total outcomes
   • = [(₄C₃)(₅C₃)] / ₉C₆
   • =[(4)(10)] / 84
   • =40/84
   • =10/21
2. P(all 5 seniors) = P(5 seniors, 1 junior) = (# ways to choose that 1 junior) / total outcomes
   • =4/84
   • =1/21
3. Since it is impossible to have a committee of 6 without including at least 2 seniors, this P = 0.

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