Question

Let 'n' be a randomly chosen integer from the set {100,101,102,.........,999}. The probability that the sum of the digits is equal to 24 is......?​

Answer 1

Answer:

answers is 1/90

bec their are only 10 no possible out of 900 numbers

Answer 2

Answer:

Concept:

Probability:

• A random event's occurrence is the subject of this area of mathematics.
• The range of the value is 0 to 1.
• Mathematics has incorporated probability to forecast the likelihood of various events.
• The degree to which something is likely to happen is basically what probability means.
• Knowing the total number of outcomes n(S) is necessary before we can calculate the likelihood that a specific event will occur.
• Let P(A) be the probability of favorable outcomes, n(A) number of favorable outcomes and n(S) be the total number of outcomes.

                                  $P(A) = \frac{n(A)}{n(S)}$

Given:

Given set is ${100,101,102,103,......999}$.

To Find:

We need to find the probability that the sum of the digits is equal to 24.

Solution:

From the set ${100,101,102,103,......999}$, we need to find the sum of the digits is 24.

The number formed when all the digits are same is $1$.

$888$

The numbers formed by 2 different digits $9,6$ is $3$.

$699, 969, 996$

The numbers formed by 3 different digits $7,8,9$ is $6$.

$789, 798, 879, 897, 978, 987$

No of the sum of the digits is equal to 24 is

$n(a) = 1+3+6 =10$

$n(S)=999-100+1=900$

⇒ $P(A) = \frac{n(A)}{n(S)}$

⇒ $P(A)=\frac{10}{900}$

⇒ $P(A)=\frac{1}{90}$

⇒ $P(A)= 0.0111$

Therefore, the probability that the sum of the digits is equal to 24 is $0.011$.

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