Math Problem Statement

Given A = {x : 10 ≤ x ≤ 21, x is an integer} and a number is chosen randomly from set A. Find the probability that: (i) the number is a multiple of 3, (ii) the sum of the two digits is an even number.

Solution

Let's go through question (c) step by step.

Given:

  • Set A={x:10x21,x is an integer}A = \{x : 10 \leq x \leq 21, x \text{ is an integer}\}.
  • We need to find the probability of two specific events if a number is chosen randomly from set AA:
  1. Event (i): The number chosen is a multiple of 3.
  2. Event (ii): The sum of the two digits of the number is even.

Step 1: Identify the Elements of Set AA

The integers between 10 and 21 are: A={10,11,12,13,14,15,16,17,18,19,20,21}A = \{10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21\} There are 12 elements in total.

Step 2: Find Probability for Event (i)

We are looking for numbers in AA that are multiples of 3. The multiples of 3 within this range are: 12,15,18,2112, 15, 18, 21 So, there are 4 numbers that meet this criterion.

The probability of Event (i) is: P(multiple of 3)=412=13P(\text{multiple of 3}) = \frac{4}{12} = \frac{1}{3}

Step 3: Find Probability for Event (ii)

For Event (ii), we need the sum of the two digits to be even. Let's calculate the digit sums for each element in AA:

  • 10:1+0=110: 1 + 0 = 1 (odd)
  • 11:1+1=211: 1 + 1 = 2 (even)
  • 12:1+2=312: 1 + 2 = 3 (odd)
  • 13:1+3=413: 1 + 3 = 4 (even)
  • 14:1+4=514: 1 + 4 = 5 (odd)
  • 15:1+5=615: 1 + 5 = 6 (even)
  • 16:1+6=716: 1 + 6 = 7 (odd)
  • 17:1+7=817: 1 + 7 = 8 (even)
  • 18:1+8=918: 1 + 8 = 9 (odd)
  • 19:1+9=1019: 1 + 9 = 10 (even)
  • 20:2+0=220: 2 + 0 = 2 (even)
  • 21:2+1=321: 2 + 1 = 3 (odd)

The numbers with an even digit sum are: 11,13,15,17,19,2011, 13, 15, 17, 19, 20. There are 6 such numbers.

The probability of Event (ii) is: P(even digit sum)=612=12P(\text{even digit sum}) = \frac{6}{12} = \frac{1}{2}

Final Answers

  1. Probability that the number is a multiple of 3: 13\frac{1}{3}.
  2. Probability that the sum of the two digits is even: 12\frac{1}{2}.

Would you like more details on any of these steps or have additional questions?

Here are some related questions to explore further:

  1. What is the probability of choosing a number that is both a multiple of 3 and has an even digit sum?
  2. How would the probabilities change if the range of xx extended from 10 to 30?
  3. Can we determine the probabilities for numbers that are multiples of other values (like 4 or 5)?
  4. What is the probability of selecting a prime number from set AA?
  5. If we chose two numbers from set AA, what is the probability both are even?

Tip: When calculating probabilities, always ensure you understand the total number of possible outcomes to set the correct denominator.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability
Number Theory
Multiples
Digit Sum

Formulas

Probability formula: P(Event) = (Number of favorable outcomes) / (Total number of outcomes)

Theorems

-

Suitable Grade Level

Grades 7-9