Chapter 5.1, Problem 27E

Triangular Numbers The first six triangular numbers (1, 3, 6, 10, 15, 21) are printed one each on one side of a card. The cards are placed face down and mixed. Choose two cards at random, and let x be the sum of the two numbers. Construct the probability distribution for this random variable x.

To determine

To construct: The probability distribution for the random variable x.

Answer to Problem 27E

The probability distribution for the random variable x is given below:

X	$P\left(X\right)$
4	$\frac{1}{15}$
7	$\frac{1}{15}$
9	$\frac{1}{15}$
11	$\frac{1}{15}$
13	$\frac{1}{15}$
16	$\frac{2}{15}$
18	$\frac{1}{15}$
21	$\frac{1}{15}$
22	$\frac{1}{15}$
24	$\frac{1}{15}$
25	$\frac{1}{15}$
27	$\frac{1}{15}$
31	$\frac{1}{15}$
36	$\frac{1}{15}$
Total	1

Explanation of Solution

Given info:

The first six numbers printed one each on side of a card is (1, 3, 6, 10, 15, 21) and x is the sum of the two numbers.

Calculation:

The first six numbers printed one each on side of a card is (1, 3, 6, 10, 15, 21). Thus the possibilities for drawing two cards is,

$\left\{\begin{array}{l}\left(1,3\right),\left(1,6\right),\left(1,10\right),\left(1,15\right),\left(1,21\right)\\ \left(3,6\right),\left(3,10\right),\left(3,15\right),\left(3,21\right)\\ \left(6,10\right),\left(6,15\right),\left(6,21\right)\\ \left(10,15\right),\left(10,21\right)\\ \left(15,21\right)\end{array}\right\}$

The total number of outcomes is 15.

The sum of two cards is ‘ $\left\{4,7,11,16,22,9,13,18,24,16,21,27,25,31,36\right\}$’.

The formula to find the probability of sum of two cards 4 is,

$P\left(\text{Sum of two cards 4}\right)=\frac{\text{Number of times having sum 4}}{\text{Total number of outcomes}}$

Substitute 1 for number of times having sum 4 and 15 for total number of outcomes,

$P\left(\text{Sum of two cards 4}\right)=\frac{\text{1}}{15}$

Similarly, the probability distribution for the random variable x is,

Sum of two cards X	Probability $P\left(X\right)$
4	$\frac{1}{15}$
7	$\frac{1}{15}$
9	$\frac{1}{15}$
11	$\frac{1}{15}$
13	$\frac{1}{15}$
16	$\frac{1}{15}+\frac{1}{15}=\frac{2}{15}$
18	$\frac{1}{15}$
21	$\frac{1}{15}$
22	$\frac{1}{15}$
24	$\frac{1}{15}$
25	$\frac{1}{15}$
27	$\frac{1}{15}$
31	$\frac{1}{15}$
36	$\frac{1}{15}$
Total	1

Justification:

Requirements for a Probability Distribution:

• The sum of probabilities of all events in the sample space must be equal to 1. That is, ${\displaystyle \sum P\left(X\right)}=1$
• The probability values of each event in the sample space must be between or equal to 0 and 1. That is, $0\le P\left(X\right)\le 1$

The random variable X represents the sum of the two numbers and these values have their corresponding probabilities. Here, the sum of all probabilities equals to 1. Also, all the individual probabilities lie between 0 and 1.