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.
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
3 answers
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).
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).
1 answer