Question

An urn contains 26 balls, each of a different color (let’s creatively call the colors c1, c2, . . . , c26). You draw four balls in sequence from the urn. Each time you draw a ball, you record the color of the ball, and return it to the urn. (That is, each time you perform this experiment, you get a sequence of four colors.)

(a) How many such four-color sequences are possible?

(b) What is the probability that a random such four-color sequence does not have any of the five colors c1, c2, c3, c4, c5?

Answer 1

(a) We need to find the number of four colour sequences possible. Note that at each step of taking out colours we have 26 choices as after taking out each ball we put it back in the urn.

So for the first position in the sequence we have 26 choices namely c1,c2,c3,...,c26. And then after noting down the color we put it back so again we have 26 choices for the second position of the sequence and so on.

So total number of possible four- colour sequences are 26 x 26 x 26 x 26 = 4,56,976

(b) Probability that a random such four-color sequence does not have any of the five colors c1,c2, c3, c4, c5. So for each position in the sequence we have 21 colour choices namely c6,c7,c8,...,c26

So the number of ways such that four-color sequence does not have any of the five colors c1,c2, c3, c4, c5 is

21 x 21 x 21 x 21 = 1,94,481

So the probability that a random such four-color sequence does not have any of the five colors c1,c2, c3, c4, c5 = number of ways such that four-color sequence does not have any of the five colors c1,c2, c3, c4, c5 divided by total number of possible four- colour sequences = 194481 / 456976 = 0.4255825251

Thank you! Please rate positively if satisfied!