We'll represent Louise's, Tammy's, Delores's, and Sheryl's point values with ,
,
, and
respectively since each one is 1 point more than the last.
Add all of these values up and set it all equal to .
Now, simplify.
Subtract on both sides.
Divide both sides by .
Since Louise's score is , the answer is
.
To verify our answer, add the point totals ,
,
, and
.
This equals , so we can be sure the answer is correct.
Answer:
a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.
b) There is a 8.65% probability that all three of the bulbs have the same rating.
c) There is a 12.45% probability that one bulb of each type is selected.
Step-by-step explanation:
There are 24 compact fluorescent lightbulbs in the box, of which:
8 are rated 13-watt
9 are rated 18-watt
7 are rated 23-watt
(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?
There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.
It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So
There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.
(b) What is the probability that all three of the bulbs have the same rating?
is the probability that all three of them are 13-watt. So:
is the probability that all three of them are 18-watt. So:
is the probability that all three of them are 23-watt. So:
There is a 8.65% probability that all three of the bulbs have the same rating.
(c) What is the probability that one bulb of each type is selected?
We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So
There is a 12.45% probability that one bulb of each type is selected.