A number x is selected at random from the numbers 1, 2, 3, 4. Another number y is selected at random from the numbers 1, 4, 9, 16. Find the probability that the product of x and y is less than 16.
Answer
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Hint: Count the number of cases present then use the formula of probability to solve the problem.
We want $x \times y < 16$ and we have 4 given values for both x and y. So, we have 16 total cases. Now, let’s count all the possible cases where $x \times y < 16$ .
$\
x \times y < 16 \\
1 \times 1 < 16 \\
1 \times 4 < 16 \\
1 \times 9 < 16 \\
2 \times 1 < 16 \\
2 \times 4 < 16 \\
3 \times 1 < 16 \\
3 \times 4 < 16 \\
4 \times 1 < 16 \\
\ $
As above mentioned, we have 8 cases which satisfy a given condition. We know the for of probability P as $p = \dfrac{{{\text{Number of favourable cases}}}}{{{\text{Number of total cases}}}}$ .On putting the values,
$p = \dfrac{8}{{{\text{16}}}} = \dfrac{1}{2}$
So $\dfrac{1}{2}$ is the required probability.
Note: there is one more method in probability theory to solve this question but we found it easier to understand.
We want $x \times y < 16$ and we have 4 given values for both x and y. So, we have 16 total cases. Now, let’s count all the possible cases where $x \times y < 16$ .
$\
x \times y < 16 \\
1 \times 1 < 16 \\
1 \times 4 < 16 \\
1 \times 9 < 16 \\
2 \times 1 < 16 \\
2 \times 4 < 16 \\
3 \times 1 < 16 \\
3 \times 4 < 16 \\
4 \times 1 < 16 \\
\ $
As above mentioned, we have 8 cases which satisfy a given condition. We know the for of probability P as $p = \dfrac{{{\text{Number of favourable cases}}}}{{{\text{Number of total cases}}}}$ .On putting the values,
$p = \dfrac{8}{{{\text{16}}}} = \dfrac{1}{2}$
So $\dfrac{1}{2}$ is the required probability.
Note: there is one more method in probability theory to solve this question but we found it easier to understand.
Last updated date: 21st Sep 2023
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