Probability: Draw Same Color/Number Chips & Find Defective Bulbs

[mehtamonica]

1. There are 5 red chips and 3 blue chips in a bowl. The red chips are numbered 1,2,3,4,5 resp and the blue chips are numbered 1,2,3 resp. If 2 chips are to be drawn at random and without replacement, find the prob that these chips have either the same number or the same color.2. In a lot of 50 light bulbs, there are 2 bad bulbs. An inspector examines 5 bulbs, which are selected at random and without replacement.

a) Find the prob of atleat one defective bulb among the 5.
b) How many bulbs should be examined so that the probability of finding atleat 1 bad bulb exceeds 1/2?

 

[pongo38]

On this forum,you have to show some initial effort before getting help. Try listing all the possible outcomes.

 

[mehtamonica]

For the first question I took 3 possible cases i) both the chips are red (ii) both the chips are blue (iii) when the no. on both the chips are the same. I added the three probabilities and got the final answer to be 4/7. I just wanted to make sure that my answer was correct.

As for the second question, the first part was simple and got the answer as 1-48C5/50C5. But the second part I was unable to solve.

 

[FourierFaux]

Things to ask yourself for the first question:

a) What is the probability of getting the combination red-red?
b) What is the probability of getting the combination blue-blue?
c) Are these probabilities additive in nature?
This tells you the probability of getting the same color. (Note: your answer to this is incorrect if you added the probability of not getting the combinations)

d) What is the probability of getting 1-1?
e) What is the probability of getting 2-2?
f) What is the probability of getting 3-3?
This tells you the probability of getting the same number. Tie it all together (Yes it's additive).

Second problem:

Part A]
Like the first ask yourself...
a) What is the probability of getting the combination b-b-g-g-g, and all it's permutations?
b) What is the probability of getting the combination b-g-g-g-g, and all it's permutations?

Part B]
Exceeds 1/2? Not "just barely exceeds" 1/2. Then your answer can be 50. ;)

On a serious aside, read this: http://www.mathwords.com/b/binomial_probability_formula.htm

Find how the formula was derived. You probably have it in your probability textbook too. :) :)


Probability theory is all about taking your problem and breaking it into a laundry list of smaller easier problems.

 

[blue_raver22]

1. The probability of drawing two chips with the same color or number can be calculated by adding the probabilities of drawing two chips with the same color and the same number, and subtracting the probability of drawing two chips with the same color and number.

The probability of drawing two chips with the same color is (5/8)*(4/7) + (3/8)*(2/7) = 5/14. This is because there are 5 red chips and 3 blue chips, and after drawing one chip, there will be 4 chips of the same color and 7 total chips remaining. The probability of drawing a second chip of the same color is then (4/7) or (2/7) depending on the color of the first chip.

The probability of drawing two chips with the same number is (5/8)*(4/7) + (3/8)*(2/7) = 5/14. This is because there are 5 chips numbered 1-5 and 3 chips numbered 1-3, and after drawing one chip, there will be 4 chips with the same number and 7 total chips remaining. The probability of drawing a second chip with the same number is then (4/7) or (2/7) depending on the number of the first chip.

The probability of drawing two chips with the same color and number is (5/8)*(4/7) + (3/8)*(2/7) = 5/14. This is because there are 5 red chips and 3 blue chips, and after drawing one chip, there will be 4 chips of the same color and 7 total chips remaining. The probability of drawing a second chip with the same color and number is then (4/7) or (2/7) depending on the color and number of the first chip.

Therefore, the overall probability of drawing two chips with the same color or number is (5/14) + (5/14) - (5/14) = 10/14 = 5/7.

2. a) The probability of selecting at least one defective bulb among 5 bulbs can be calculated by subtracting the probability of selecting 5 good bulbs from 1.

The probability of selecting 5 good bulbs is (48/50)*(47/49)*(46/48)*(45/47)*(44/46) = 0