Question

(No work needed) Jar A contains the 8 letters in the word Colorado. Jar B contains the 11 letters in the word Connecticut. What is the probability of randomly drawing an (n) or a (t) from Jar B

Answer 1

Jar B contains 11 letters, including the letters in the word "Connecticut". To find the probability of randomly drawing an (n) or a (t) from Jar B, we need to calculate the total number of (n)s and (t)s in Jar B and divide that by the total number of letters in the jar.

There are 3 (n)s and 2 (t)s in the word "Connecticut". Therefore, there are a total of 3 + 2 = 5 letters in Jar B that are either (n)s or (t)s.

The total number of letters in Jar B is 11.

Therefore, the probability of randomly drawing an (n) or a (t) from Jar B is:

P = (number of (n)s and (t)s in Jar B) / (total number of letters in Jar B)

P = 5/11

So the probability of randomly drawing an (n) or a (t) from Jar B is 5/11 or approximately 0.45 (rounded to two decimal places).

Answer 2

Explanation:

The probability of drawing an "n" from Jar B is 2/11, since there are two "n"s out of a total of 11 letters. Once we draw an "n", there are 10 letters left in the jar, including one "t". Since we're not replacing the "n" that we drew, the probability of drawing a "t" next is 1/10.

Therefore, the probability of drawing an "n" from Jar B and then drawing a "t" without replacement is:

(2/11) * (1/10) = 2/110 = 1/55

So the probability is 1/55.