A bag contains 11 oranges, four of which are damaged. Two oranges are selected at random without replacement. The probability that both oranges are not damaged is:
To find the probability that both oranges are not damaged, we need to calculate the probability of selecting one undamaged orange and then selecting another undamaged orange from the remaining oranges.
There are 11 oranges in total, 4 of which are damaged. So, there are 11 - 4 = <<11-4=7>>7 undamaged oranges.
The probability of selecting an undamaged orange on the first draw is 7/11.
Once one undamaged orange is selected, there are 10 remaining oranges in total, including 6 undamaged oranges. So, the probability of selecting an undamaged orange on the second draw is 6/10.
To find the probability of both events happening, we multiply the probabilities: (7/11) * (6/10) = 42/110.
Therefore, the probability that both oranges are not damaged is 42/110 or 0.3818 (rounded to four decimal places).
To calculate the probability that both oranges are not damaged, we need to determine the number of ways to select two undamaged oranges out of the total number of ways to select any two oranges.
Step 1: Calculate the total number of ways to select any two oranges from the bag.
Since there are 11 oranges in total, we can use the combination formula to calculate the total number of ways to select any two oranges:
C(11, 2) = (11!)/(2!(11-2)!) = (11!)/(2!9!) = (11x10)/(2x1) = 55.
Step 2: Calculate the number of ways to select two undamaged oranges.
Since 4 oranges are damaged, there are 11-4= 7 undamaged oranges in the bag. We can apply the combination formula again to calculate the number of ways to select any two undamaged oranges:
C(7, 2) = (7!)/(2!(7-2)!) = (7!)/(2!5!) = (7x6)/(2x1) = 21.
Step 3: Calculate the probability that both selected oranges are undamaged.
The probability is calculated by dividing the number of favorable outcomes (selecting two undamaged oranges) by the total number of possible outcomes (selecting any two oranges):
P(both undamaged) = (number of ways to select two undamaged oranges)/(total number of ways to select any two oranges) = 21/55 ≈ 0.381.
Therefore, the probability that both oranges are not damaged is approximately 0.381 or 38.1%.