Abox in a certain supply room contains four 40w, five 60w, and six 75w light-bulbs. suppose that three bulbs are randomly selected. a. what is the probability that exactly two of the selected bulbs are rated 75w? 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 75w bulb is found. what is the probability that it is necessary to examine at least six bulbs?

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

[tex]K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455[/tex]

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

[tex]\frac{270}{455}=0.5934=59.34\%[/tex]

(b)

The probability of selecting three 40-W bulbs is

[tex]\frac{4*3*2}{455}=0.0527=5.27\%[/tex]

The probability of selecting three 60-W bulbs is

[tex]\frac{5*4*3}{455}=0.1318=13.18\%[/tex]

The probability of selecting three 75-W bulbs is

[tex]\frac{6*5*4}{455}=0.2637=26.37\%[/tex]

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

[tex]\frac{6*5*4}{455}=0.2637=26.37\%[/tex]

(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 [tex]\frac{9}{15}=0.6[/tex]

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

Following this procedure 5 times, we find the probabilities

[tex]\frac{9}{15},\frac{8}{14},\frac{7}{13},\frac{6}{12},\frac{5}{11}[/tex]

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%

Situation satisfies the criteria for the use of hypergeometric distribution. Since no replacement is made, binomial distribution is not applicable (probability does not remain constant).

A=number of target wattage bulbs

B=number of non-targeted wattage bulbs

a=number of target wattage bulbs selected

b=number of non-targeted wattage bulbs selected

P(a,b)=C(A,a)*C(B,b)/C(A+B,a+b)

where C(n,x)=combination of x items chosen from n=n!/(x!(n-x)!)

For all following problems,

A+B=4+5+6=15

a+b=3 (selected)

(a) Target wattage = 75W

A=6, B=9, a=2, b=1

P(a,b,A,B)

=P(2,1,6,9)

=C(6,2)*C(9,1)/C(15,3)

=15*9/455

=27/91

(b) target wattage = each of the three

Probability = sum of probabilities of choosing 3 40,60,75-watt bulbs

P(3x40W)+P(3x60W)+P(3x75W)

Case (A,B,a,b)

3x40W (4,11,3,0)

3x60W (5,10,3,0)

3x75W(6,9,3,0)

P(3x40W)+P(3x60W)+P(3x75W)

=C(4,3)*C(11,0)/C(15,3)+C(5,3)*C(10,0)/C(15,3)+C(6,3)*C(9,0)/C(15,3)

=4*1/455+10*1/455+20*1/455

=34/455

Can also be solved by elementary counting, for example, for (a),

P(2x75W)

=C(3,2)*6/15*5/14*9/13

=(3)*6/15*5/14*9/13

=27/91 as before

Probability that exactly three of the selected bulbs are rated 75-W is 0.044.

Step-by-step explanation:

We are given that a box in a certain supply room contains four 40-W light bulbs, five 60-W bulbs, and six 75-W bulbs.

And we have to find the probability that exactly three of the selected bulbs are rated 75-W.

Here, we will use Combinations for selection procedure, i.e.;

[tex]^{n} C_r = \frac{n!}{r! \times (n-r) !}[/tex]

So, number of ways of selecting three 75-W rated bulbs from total of six 75-W rated bulbs in a box is given by = [tex]^{6} C_3[/tex]

Total number of ways of selecting three bulbs from total of 15 bulbs (4 + 5 + 6) in a box = [tex]^{15} C_3[/tex]

Therefore, probability that exactly three of the selected bulbs are rated 75-W is given as = [tex]\frac{^{6} C_3}{^{15} C_3}[/tex] = [tex]\frac{\frac{6!}{3! \times 3!} }{\frac{15!}{3! \times 12!} }[/tex]

= [tex]\frac{6!}{3!} \times \frac{12!}{15!}[/tex] = [tex]120 \times \frac{1}{2730}[/tex] = [tex]\frac{4}{91}[/tex]

= 0.044

Hence, the required probability is 0.044.

a) There are 126 different ways are there to select four of the nine computers to be set up.

b) 31.75% probability that exactly three of the selected computers are desktops.

c) 35.71% probability that at least three desktops are selected.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem, there are no replacements. Which means that after one of the 9 computers is selected, there will be 8 computers.

Also, the order that the computers are selected is not important. For example, desktop A and desktop B is the same outcome as desktop B and desktop A.

These are the two reasons why the combinations formula is important to solve this problem.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

(a) How many different ways are there to select four of the nine computers to be set up?

Four computers are selected from a set of 9.

So

[tex]T = C_{9,4} = \frac{9!}{4!(9-4)!} = 126[/tex]

There are 126 different ways are there to select four of the nine computers to be set up.

(b) What is the probability that exactly three of the selected computers are desktops?

Desired outcomes:

3 desktops, from a set of 5

One laptop, from a set of 4.

So

[tex]D = C_{5,3}*C_{4,1} = 40[/tex]

Total outcomes:

From a), 126

Probability:

[tex]P = \frac{40}{126} = 0.3175[/tex]

31.75% probability that exactly three of the selected computers are desktops.

(c) What is the probability that at least three desktops are selected?

Three or four

Three:

[tex]P = \frac{40}{126}[/tex]

Four:

Desired outcomes:

4 desktops, from a set of 5

Zero laptop, from a set of 4.

[tex]D = C_{5,4}*C_{4,0} = 5[/tex]

[tex]P = \frac{5}{126}[/tex]

Total(three or four) probability:

[tex]P = \frac{40}{126} + \frac{5}{126} = \frac{45}{126} = 0.3571[/tex]

35.71% probability that at least three desktops are selected.

Yesterday the bandits scored 56 points in their basketball game. today they scored 42 points. what is the percent decreasw of the bandits score from yesterday to today?

Kindly check explanation

Step-by-step explanation:

Given the following :

Number of white balls = 5

Number of red balls = 4

Number if blue balls = 3

Probability = required outcome / Total possible outcomes

Total possible outcomes (Total number of balls) = 5 + 4 + 3 = 12

a. What is the probability that exactly two of the balls are white?

(White, white, not white) or (not white, white, white) or (white, not white, white)

Assume balls were picked without replacement ;

P(white, white, not white) = (5/12 × 4/11 × 7/10) = 0.1060606

P(not white, white, white) = (7/12 × 5/11 × 4/10) = 0.1060606

P(white, not white, white) = (5/12 × 7/11 × 4/10) = 0.1060606

P(exactly 2 white balls) = 3×(0.1060606) = 0.3181818

b. What is the probability that one ball of each color is selected ?

(White, red, blue) or (red, white, blue) or (red, blue, white) or (blue, red, white), (white, blue, red), (blue, white, red)

P(white, red, blue) = (5/12 * 4/11 * 3/10) = 0.0454545

(0.0454545 × 3) = 0.272727

c. What is the probability that all three of the selected balls have the same color ?

P(white, white, white) or P(blue, blue, blue) or P(red, red, red)

P(white, white, white) = (5/12 * 4/11 * 3/10) = 0.0454545

P(blue, blue, blue) = (3/12 * 2/11 * 1/10) = 0.0045454

P(red, red, red) = (4/12 * 3/11 * 2/10) = 0.0181818

(0.0454545 + 0.0045454 + 0.0181818) = 0.0681817

answer:

b) 31.75% probability that exactly three of the selected computers are desktops.

c) 35.71% probability that at least three desktops are selected.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.In this problem, there are no replacements. Which means that after one of the 9 computers is selected, there will be 8 computers.Also, the order that the computers are selected is not important. For example, desktop A and desktop B is the same outcome as desktop B and desktop A.These are the two reasons why the combinations formula is important to solve this problem.

⭐ Answered by Kakashi ʕ •㉨• ʔ⭐

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answer:the answer is B

Step-by-step explanation: