Question

An urn contains 3 yellow chips, numbered 1 through 3, four red chips, numbered 1 through 4, and 5 green chips, numbered 1 through 5.An urn contains 3 yellow chips, numbered 1 through 3, four red chips, numbered 1 through 4, and 5 green chips, numbered 1 through 5.

1. What is the probability of not drawing a chip numbered 4 on the draw of a single chip?
2. What is the probability of drawing a chip numbered 3 or a green chip on the draw of a single chip?
3. What is the probability of drawing two yellow chips (of any numbers) when drawing two chips and not replacing the first chip drawn?
4. What is the probability of drawing two yellow chips (of any numbers) when drawing two chips and replacing the first chip drawn?
5. What is the probability of getting at least one chip numbered 1 when drawing two chips and not replacing the first chip drawn?

Answer 1

Total number of chips in the urn = 12 (3 yellow , 4 red , 5 green)

Answer to 1st question :
Number of 4 numbered chips in the urn = 2 (1 red and 1 green)

and , number of chips that are not 4 numbered = 12-2 = 10

Therefore ,

Probability of not drawing a chip numbered 4 on the draw of a single chip = 10C1 / 12C1 = 10/12 = 5/6 = 0.8333

Answer to 2nd question :

Number of green chips in the urn = 5

Probability of drawing a green chip on the draw of a single chip , P(G) = 5C1 / 12C1 = 5/12

Number of 3 numbered chips in the urn = 3 (1 green , 1 red , 1 yellow)

Probability of drawing a chip numbered 3 on the draw of a single chip , P(3c) = 3C1 / 12C1 = 3/12

Probability of drawing a 3 numbered green chip , P(3cG) = 1/12

Probability of drawing a chip numbered 3 or a green chip on the draw of a single chip = P(G) + P(3c) - P(3cG)

= (5/12) + (3/12) - (1/12) = 7/12 = 0.5833

Answer to 3rd question :
Probability of drawing a yellow chip in the first draw = 3C1 / 12C1 = 3/12

Probability of drawing a yellow chip in the second draw = 2C1 / 11C1 = 2/11 (Since , the first chip was not replaced)

Therefore ,

Total probability of drawing two yellow chips (of any numbers) when drawing two chips and not replacing the first chip drawn = (3/12) x (2/11) = 0.0455

Answer to 4th question :

Probability of drawing a yellow chip in the first draw = 3C1 / 12C1 = 3/12

Probability of drawing a yellow chip in the second draw = 3C1 / 12C1 = 3/12 (Since , the first chip was replaced)

Therefore ,

Total probability of drawing two yellow chips (of any numbers) when drawing two chips and replacing the first chip drawn = (3/12) x (3/12) = 0.0625

Answer to 5th question :
Number of 1 numbered chips in the urn = 3 (1 yellow , 1 red , 1 green)

and , number of chips that are bot 1-numbered = 9

There are 2 cases

• Case I : drawing 1 1-numbered chip and the 2nd chip of any number

Probability of getting 1-numbered chip and the 2nd chip of any number = (3C1 / 12C1) x (9C1 / 11C1) = 0.2045

• Case II : drawing 2 1-numbered chips

Probability of drawing two 1-numbered chips = (3C1 / 12C1 ) x (2C1 / 11C1) = 0.0455

(The chips are drawn using WITHOUT REPLACEMENT procedure)

Therefore , total probability of getting at least one chip numbered 1 when drawing two chips and not replacing the first chip drawn = 0.0455 + 0.2045 = 0.25

(NOTE THAT : In case I , I didn't consider the order in which the chips are drawn , if ordering is a matter of concern then , the probability of 1st case changes to 0.409 and the total probability equals 0.6135)