Question

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A number 'x' is selected from the numbers 1, 2, 3 and then a second 'y' is randomly selected from the numbers 1,4,9 . What is the probability that the product 'xy' of the two numbers will be less than 9?
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Answer 1

Number x can be selected in three ways and corresponding to each such way there are three ways of selecting number y. Therefore, two numbers can be selected in 9 ways as listed :

(1,1),(1,4),(1,9),(2,1),(2,4),(2,9),(3,1),(3,4),(3,9)

So, total number of elementary events = 9

The product xy will be less than 9, if x, and y are chosen in one of the following ways :

(1,1),(1,4),(2,1),(2,4),(3,1)

Therefore, favourable number of elementary events = 5

Therefore, required probability = 95

Answer 2

Given :-

$\text{A number x is selected from the numbers 1,2,3}$

$\text{ and then a second number y is}$ $[tex]\text{randomly selected from the numbers 1,4,9.l[tex]</p><p></p><p>[tex] \text{A number x is selected from the numbers 1,2,3} [tex] \text{ and then a second number y is}[/tex </p><p></p><p>[tex] \text{randomly selected from the numbers 1,4,9.}$

$\text{Total \: number \: of \: outcomes \: for \: product } \\ \text{of \: two \: numbers \: selected \: from \: x \: any :}$

$\boxed{\text{3×3=9}}$

$\text{The outcomes for which the product xy of the} \\ \text{two numbers selected will be less than 9 : } \\ \text{1×1 , 1×4, 2×1, 2×4, 3×1}$

$\text{Hence, the number of desired outcomes = 5}$

$\text{Then, the probability that the product xy of the} \\ \text{two numbers selected will be less than 9 :-}$

$= \sf\dfrac {Desired \: out \: come}{Total \: outcome}$

$= \sf\dfrac{5}{9}$