A bag contains 11 oranges, four of which are damaged. Two oranges are selected at random without replacement. The probability that both oranges are damaged is
To find the probability that both oranges are damaged, we need to consider the probability of selecting a damaged orange on the first pick and a damaged orange on the second pick.
On the first pick, there are 11 oranges to choose from, and 4 of them are damaged. So, the probability of selecting a damaged orange on the first pick is 4/11.
After one orange has been picked, there are 10 oranges left in the bag, and 3 damaged oranges left. So, the probability of selecting a damaged orange on the second pick, without replacement, is 3/10.
To find the probability that both oranges are damaged, we multiply these probabilities together:
(4/11) * (3/10) = 12/110 = 6/55
The probability that both oranges are damaged is 6/55.
Assume X to be the number of heads obtained in 70 independent tosses of a balance coin. Therefore, X is a Binomial random variable with
X is a binomial random variable with parameters n = 70 and p = 0.5.
In this case, the number of independent tosses is 70, and each toss has a probability of success (getting a head) denoted by p = 0.5 (assuming a balanced coin).
Therefore, X follows a binomial distribution with n = 70 (the number of trials) and p = 0.5 (the probability of success in each trial).
To find the probability that both oranges are damaged, we need to calculate the probability of selecting a damaged orange for the first pick, and then the probability of selecting another damaged orange on the second pick.
First, let's calculate the probability of selecting a damaged orange for the first pick. There are a total of 11 oranges, and 4 of them are damaged. So, the probability of picking a damaged orange on the first pick is 4/11.
Next, since we are picking without replacement, there will be one less orange in the bag for the second pick. After the first pick, there are 10 oranges in the bag, and 3 of them are damaged. So, the probability of picking a damaged orange on the second pick is 3/10.
To calculate the probability of both events happening, we multiply the probabilities together:
P(both oranges are damaged) = (4/11) * (3/10)
Simplifying the expression, we get:
P(both oranges are damaged) = 12/110
Therefore, the probability that both oranges are damaged is 12/110, which can be simplified further if needed.