A bucket contains 7 red, 3 blue, 2 black and 1 yellow ball of the same size.

A random selection of 5 balls is taken from the bucket.
find the probability that the random selection will contain atleast 1 red ball?

pr(1 or more red)=1-pr(no red)

= 1-(6/13)(5/12)(4/11)(3/10)(2/9)

To find the probability of selecting at least 1 red ball, we need to calculate the probability of selecting 1 red ball, 2 red balls, 3 red balls, 4 red balls, and 5 red balls, and then add them all together.

Let's start by calculating the probability of selecting 1 red ball. There are a total of 13 balls in the bucket, and 7 of them are red. So the probability of selecting 1 red ball is:

P(1 red ball) = (Number of ways to select 1 red ball) / (Total number of possible selections of 5 balls)

The number of ways to select 1 red ball out of 7 red balls is given by the combination formula:

Number of ways to select 1 red ball = C(7, 1)

where C(n, r) represents the number of combinations of n items taken r at a time.

The total number of possible selections of 5 balls out of 13 is given by:

Total number of possible selections = C(13, 5)

Now we can calculate the probability of selecting 1 red ball:

P(1 red ball) = C(7, 1) / C(13, 5)

Similarly, we can calculate the probabilities of selecting 2, 3, 4, and 5 red balls using the same approach:

P(2 red balls) = C(7, 2) / C(13, 5)
P(3 red balls) = C(7, 3) / C(13, 5)
P(4 red balls) = C(7, 4) / C(13, 5)
P(5 red balls) = C(7, 5) / C(13, 5)

Finally, we can find the probability of selecting at least 1 red ball by adding up these probabilities:

P(at least 1 red ball) = P(1 red ball) + P(2 red balls) + P(3 red balls) + P(4 red balls) + P(5 red balls)

Note: The combination formula can be calculated using the factorial function. C(n, r) = n! / (r! * (n-r)!)