Question

. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected. a. What is the probability that exactly two of the selected bulbs are rated 75-W? b. What is the probability that all three of the selected bulbs have the same rating? c. What is the probability that one bulb of each type is selected? d. Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?

Answer 1

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total 4+5+6 = 15 bulbs. If we want to select 3 randomly there are  K ways of doing this, where K is the combination of 15 elements taken 3 at a time

$K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455$

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

$\frac{270}{455}=0.5934=59.34\%$

(b)

The probability of selecting three 40-W bulbs is

$\frac{4*3*2}{455}=0.0527=5.27\%$

The probability of selecting three 60-W bulbs is

$\frac{5*4*3}{455}=0.1318=13.18\%$

The probability of selecting three 75-W bulbs is

$\frac{6*5*4}{455}=0.2637=26.37\%$

Since the events are disjoint, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

$\frac{6*5*4}{455}=0.2637=26.37\%$

(d)

The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, supposing there is no replacement, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is $\frac{9}{15}=0.6$

Since there are no replacement, the probability of taking a second non 75-W bulb is now $\frac{8}{14}=0.5714$

Following this procedure 5 times, we find the probabilities

$\frac{9}{15},\frac{8}{14},\frac{7}{13},\frac{6}{12},\frac{5}{11}$

which are

0.6, 0.5714, 0.5384, 0.5, 0.4545

As the events are independent, the probability of choosing 5 non 75-W bulbs is the product

0.6*0.5714*0.5384*0.5*0.4545 = 0.0419 = 4.19%

Answer 2

Probabilities are used to determine the possibility of an event. The following are the probabilities of the events.

• The probability that two of the selected bulbs are 75W is 0.2967
• The probability that all the selected bulbs are the same is 0.0747
• The probability that one of each type are selected is 0.2637
• The probability that at least six bulbs is selected until 75W bulb is found is 0.1591

The total number of bulbs is:

$Total = 4 + 5 + 6 = 15$

First, we calculate the number of ways of selecting 3 bulbs from the 15 bulbs.

There are: $^{15}C_3$ ways to select 3 bulbs from the 15.

$^{15}C_3 = 455$

(a): Probability that two of the selected bulbs are 75W

If two 75W bulbs are selected, then one non-75W is selected.

The number of ways of selecting two from six 75W bulbs is:

$^6C_2 = \frac{6!}{4!2!}$

$^6C_2 = 15$

The number of ways of selecting one from nine non 75W bulbs is:

$^9C_1 = \frac{9!}{8!1!}$

$^9C_1 = 9$

So, the total selection is:

$Total = 15 \times 9 = 135$

So, the probability that two of the selected bulbs are 75W is:

$Pr = \frac{135}{455}$

$Pr = 0.2967$

Hence, the probability that two of the selected bulbs are 75W is 0.2967

(b): Probability that all the selected bulbs are the same

The number of ways of selecting three of four 40W bulbs is:

$^4C_3 = \frac{4!}{3!1!}$

$^4C_3 = 4$

The number of ways of selecting three of five 60W bulbs is:

$^5C_3 = \frac{5!}{3!2!}$

$^5C_3 = 10$

The number of ways of selecting three of six 75W bulbs is:

$^6C_3 = \frac{6!}{3!3!}$

$^6C_3 = 20$

The total selection is:

$Total = 4 + 10 + 20 = 34$

So, the probability that all the selected bulbs are the same is:

$Pr = \frac{34}{455}$

$Pr = 0.0747$

Hence, the probability that all the selected bulbs are the same is 0.0747

(c): Probability that one of each type are selected

The number of ways of selecting one of four 40W bulbs is:

$^4C_1 = \frac{4!}{3!1!}$

$^4C_1 = 4$

The number of ways of selecting one of five 60W bulbs is:

$^5C_1 = \frac{5!}{4!1!}$

$^5C_1 = 5$

The number of ways of selecting three of six 75W bulbs is:

$^6C_1 = \frac{6!}{5!1!}$

$^6C_3 = 6$

The total selection is:

$Total = 4 \times 5 \times 6 = 120$

So, the probability that one of each type are selected

$Pr = \frac{120}{455}$

$Pr = 0.2637$

Hence, the probability that one of each type are selected is 0.2637

(d): Probability that at least six bulbs is selected until 75W bulb is found

This means that the first five are non 75W bulbs.

There are nine non 75W bulbs.

The number of ways of selecting five of nine none 75W bulbs is:

$^9C_5 = \frac{9!}{4!5!}$

$^9C_5 = 126$

The number of ways of selecting the first five bulbs from the 15 bulbs is:

$^{12}C_5 = \frac{12!}{7!5!}$

$^{12}C_5= 792$

So, the probability that at least six bulbs is selected until 75W bulb is found is:

$Pr = \frac{126}{792}$

$Pr = 0.1591$

Hence, the probability that at least six bulbs is selected until 75W bulb is found is 0.1591

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