A newspaper company is selecting four houses to receive a free newspaper. There are 10 houses on your block that are numbered 1-10. What is the probability that the four houses selected will all be even numbered houses? *
To solve this problem, we need to determine the total number of possible outcomes and the number of favorable outcomes.
Total number of possible outcomes:
There are 10 houses in total, so the number of possible outcomes is 10C4 (read as 10 choose 4), which can be calculated using combinations formula:
nCr = n! / (r!(n-r)!)
Here, n is the total number of houses (10) and r is the number of houses to be selected (4).
10C4 = 10! / (4!(10-4)!)
= 10! / (4!6!)
Number of favorable outcomes:
Since we want to select four houses that are all even numbered, we need to determine the number of even numbered houses available:
Out of the 10 houses, 5 are even numbered (2, 4, 6, 8, and 10). Therefore, the number of favorable outcomes is 5C4.
5C4 = 5! / (4!(5-4)!)
= 5! / (4!1!)
Now, we can easily calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = 5C4 / 10C4
Let's calculate these values to find the probability.
To find the probability that the four houses selected will all be even numbered houses, we need to determine the total number of possible outcomes and the number of favorable outcomes.
Total number of possible outcomes: Since there are 10 houses on the block and the newspaper company is selecting four houses, the total number of possible outcomes is determined by the combination formula:
C(n, r) = n! / (r!(n-r)!)
Where n is the total number of items and r is the number of items selected.
In this case, n = 10 and r = 4.
C(10, 4) = 10! / (4!(10-4)!)
= 10! / (4!6!)
= (10 x 9 x 8 x 7) / (4 x 3 x 2 x 1)
= 210
Therefore, there are 210 possible outcomes.
Number of favorable outcomes: Since we want all four houses selected to be even numbered, there are a total of 5 even numbered houses on the block (2, 4, 6, 8, and 10). So, we need to select 4 even numbered houses from these 5 houses.
C(5, 4) = 5! / (4!(5-4)!)
= 5! / (4!1!)
= 5
Therefore, there are 5 favorable outcomes.
Probability that all four houses selected will be even numbered houses:
Probability = Number of favorable outcomes / Total number of possible outcomes
P = 5 / 210
P ≈ 0.0238
Therefore, the probability that the four houses selected will all be even numbered houses is approximately 0.0238 or 2.38%.
2 4 6 8 10
first one 5/10
second one 4/9
third one 3/8
fourth one 2/7
multiply