Let $A = \left\{ 1,\ 3,\ 5,\ 7,\ 9 ight\},\ B = \left\{ 2,\ 4,\ 6,\ 8 ight\}.$ If a cartesian product $A imes B$, if chosen at random, the probability of $a + b = 9$ is

Question

Let $A = \left\{ 1,\ 3,\ 5,\ 7,\ 9 ight\},\ B = \left\{ 2,\ 4,\ 6,\ 8 ight\}.$ If a cartesian product $A 	imes B$, if chosen at random, the probability of $a + b = 9$ is

Dr. Abubacker · Accepted Answer

(b)
Total number of cases=20
Favorable cases=$\left\{ (1,8),(3,6),(5,4),(7,2) ight\} = 4$
$	herefore$ Required probability $= \frac{4}{20} = \frac{1}{5}$

Q. Let $A = \left\{ 1,\ 3,\ 5,\ 7,\ 9 \right\},\ B = \left\{ 2,\ 4,\ 6,\ 8 \right\}.$ If a cartesian product $A \times B$, if chosen at random, the probability of $a + b = 9$ is

Explanation

Reviewed by Dr. Abubacker

Want more practice?