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# Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking into it. What is the probability of?(i).Getting a number 6?(ii).Getting a number less than 6?(iii).Getting a number greater than 6?(iv).Getting a 1 digit number? Verified
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Hint: Use the basic definition of probability. First find out the number of cases which are favorable to the given condition in each case and then find out the total number of possible outcomes. And finally find out the ratio.

Given that there are 10 unique slips with different numbers on it.
So total number possible outcomes = 10.

(i).Getting a number 6
Among all the given numbers favorable outcome is $\left\{ 6 \right\}$
Number of favorable outcome $= 1$
Total number of outcomes $= 10$
Probability of getting a number 6
$= \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}} \\ = \dfrac{1}{{10}} \\$
Hence, the probability of getting a number 6 is $\dfrac{1}{{10}}$ .

(ii).Getting a number less than 6
Among all the given numbers favorable outcome is $\left\{ {1,2,3,4,5} \right\}$
Number of favorable outcome $= 5$
Total number of outcomes $= 10$
Probability of getting a number less than 6
$= \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}} \\ = \dfrac{5}{{10}} \\ = \dfrac{1}{2} \\$
Hence, the probability of getting a number less than 6 is $\dfrac{1}{2}$ .

(iii).Getting a number more than 6
Among all the given numbers favorable outcome is $\left\{ {7,8,9,10} \right\}$
Number of favorable outcome $= 4$
Total number of outcomes $= 10$
Probability of getting a number more than 6
$= \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}} \\ = \dfrac{4}{{10}} \\ = \dfrac{2}{5} \\$
Hence, the probability of getting a number more than 6 is $\dfrac{2}{5}$ .

(iv).Getting a 1 digit number
Among all the given numbers favorable outcome is $\left\{ {1,2,3,4,5,6,7,8,9} \right\}$
Number of favorable outcome $= 9$
Total number of outcomes $= 10$
Probability of getting a 1 digit number
$= \dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total number of outcomes}}}} \\ = \dfrac{9}{{10}} \\$
Hence, the probability of getting a 1 digit number is $\dfrac{9}{{10}}$ .

Note: We know that probability is defined as the ratio of number of favorable outcomes of an event to the total number of outcomes of the event. In the above question, we just found out the favorable outcomes of the event and total number of outcomes. Probability of an event will always be between 0 and 1.
Last updated date: 20th Sep 2023
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