If numbers 1-10 are put in a bowl determine the probability that if two are selected they would 1. both be even,2. both would be less than 5 and 3. the first number is odd and the second number is even.
both even= 5/10 * 4/9
Whate is the answer
To determine the probabilities, we need to know the total number of possible outcomes (sample space) and the number of favorable outcomes (events that satisfy the given condition). Let's analyze each case:
1. Both numbers are even:
The even numbers between 1 and 10 are 2, 4, 6, 8, 10. So, there are 5 even numbers in the bowl. To find the probability of selecting two even numbers, we need to calculate the number of ways to choose 2 even numbers out of 5 and divide it by the total number of possible outcomes.
Total possible outcomes: The total number of balls in the bowl is 10, and we are choosing 2 numbers. Thus, we have to calculate the number of combinations of 2 out of 10 balls: C(10, 2) = 10! / (2! * (10-2)!) = 45.
Favorable outcomes: We need to choose 2 even numbers out of the 5 even numbers: C(5, 2) = 5! / (2! * (5-2)!) = 10.
The probability of selecting two even numbers = favorable outcomes / total possible outcomes = 10 / 45 ≈ 0.222 (or 22.2%).
2. Both numbers are less than 5:
The numbers less than 5 in the bowl are 1, 2, 3, 4. So, there are 4 numbers that satisfy the condition. Similar to the previous case, we need to calculate the number of ways to choose 2 numbers out of 4 and divide it by the total number of possible outcomes.
Total possible outcomes: We still have a total of 10 balls, but this time we are only choosing 2 numbers. So, C(10, 2) = 45.
Favorable outcomes: We need to choose 2 numbers out of the 4 satisfying the condition: C(4, 2) = 4! / (2! * (4-2)!) = 6.
The probability of selecting two numbers less than 5 = favorable outcomes / total possible outcomes = 6 / 45 ≈ 0.133 (or 13.3%).
3. The first number is odd, and the second number is even:
To calculate this probability, we need to consider two sub-cases:
a) The first number is odd: There are 5 odd numbers in the bowl.
b) The second number is even: There are 5 even numbers remaining after removing the odd number chosen for the first position.
Total possible outcomes: Just like before, C(10, 2) = 45.
Favorable outcomes: We need to choose 1 odd number and 1 even number: C(5, 1) * C(5, 1) = 5 * 5 = 25.
The probability of selecting an odd number followed by an even number = favorable outcomes / total possible outcomes = 25 / 45 ≈ 0.556 (or 55.6%).
Remember, the probabilities are approximate because they involve division and rounding.