Question

A package contains 4 light bulbs. The following table gives the probability distribution (pdf) of the number of broken bulbs in the package. A random package is selected. Use the table to answer the following questions:

pdf for DRV

X |
0 |
1 |
2 |
3 |
4 |

p(x) |
0.66 |
0.20 |
? |
0.05 |
0.02 |

- What is the probability that there are exactly 2 defective bulbs in a package?
- What is the probability that there are at most 2 defective bulbs in a package?
- What is the expected number (mean) of defective bulbs in a package?
- Find the standard deviation of the number of defective bulbs in a package.

If two independent packages are selected, what is the probability that there is at least 1 total defective bulb?

Answer #1

1) We know that the probability always lies between O to 1.

And pi = 1

So p(x=0) + p(x=1) +p(x=2) + p(x=3) + p(x=4) =1

P(x=2) + 0.660 + 0.20+0.05 + 0.02 =1

P(x =2) = 1-0.93

P(x =2) = 0.07

The probability of Exactly two defective bulbs is 0.07.

2) p(x2) = p(x=0) + p(x=1) + p(x=2)

= 0.660 + 0.20 + 0.07

P(x 2) = 0.93

3) expected number

x*p(x) = 0*0.66 + 1*0.20 + 2*0.07 + 3*0.05 + 4*0.02

= 0 + 0.20 + 0.14 + 0.15 + 0.08

Expected number = 0.57

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