Question

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$\large\purple{\underline{{\boxed{\textbf{Question:-}}}}}$

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

$\sf\underline{\bigstar\:\sf Answer}$

Total number of ways the two numbers ca be selected is :

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

So the total number of outcomes is 9

ATQ, the product must not exceed more than 9

Hence,

Possible out comes are:

➳ (1,1)

➳ (1,4)

➳ (2,1)

➳ (2,4)

➳ (3,1)

So the possible number of outcomes is 5

So the probably will be given by,

»» $\displaystyle\sf Probability = \dfrac{Possible \ outcomes}{Total \ no \ of \ outcomes}$

◕ Possible Outcomes = 5

◕ Total outcomes = 9

»» $\displaystyle\sf \pink{Probability = \dfrac{5}{9}}$

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Answer 2

Given:

• A number 'x' is randomly selected from the numbers 1, 2, 3
• Then, number 'y' is randomly selected from the numbers 1,4,9

Find:

• Probability that the product 'xy' of two numbers will be less that 9

Solution:

Here, number 'x' can be selected in three ways and corresponding to each such way

and the number 'y' can also be selected in three ways

Therefore, two numbers can be selected in the 9 ways as listed below:

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

➟So, total number of elementry events = 9

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

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

➟So, favourable number of elementry events = 5

$\sf \to Probability_{(E)}= \dfrac{Favourable \: Outcomes}{Total \: Outcomes}$

where,

❂Favourable Outcomes = 5

❂Total Outcomes = 9

So,

$\sf \to Probability_{(xy < 9)}= \dfrac{Favourable \: Outcomes}{Total \: Outcomes}$

$\sf \to Probability_{(xy < 9)}= \dfrac{5}{9}$

___________________

Hence, Required Probability is 5/9