Question

Three defective electric toothbrushes were accidentally shipped to a drugstore by Fresh
and White products along with 17 nondefective ones.
i What is the probability the first two electric toothbrushes sold will returned to the
drugstore because they are defective?
ii What is the probability the first two electric toothbrushes sold will not be defective?

Answer 1

Probability of 1 defective sold = Pd1 = 3/20
3 defective out of 20 total

Probability of 2 defective sold =Pd2= (3-1)/(20-1) = 2/19
one has already been sold ( 3-1), (20-1)

Probability of them both being sold consecutively = Pd1*Pd2 = 3/20 * 2/19

For the second part, you use 17 instead of 3 to start out.

Hope this helps :)

Answer 2

c. If you ask three strangers on campus, what is the probability:
i All were born on Wednesday?
ii All were born different days of the week?
iii None were born on Saturday?

Answer 3

I) The probability that

1 person is born on a Wednesday is

1/7 (one day out of 7 possible),

that 2 are both born on a Weds is

1/7*1/7 (the second person also has 1 out 7 possible days)

so three on the same day would be.....


ii) The probability that a person is born on a day is

1, they were definitely born on a day

the probability that another person is born on a different day means

there are only 6 days left to choose from, so

1 * 6/7.

The third person only has 5 choices left, so the answer would be....

iii) This one is like the first one. The probability of 1 person not being born on a Saturday is

6/7 - there are 6 days the person can be born on that aren't Saturday.

Probability that two people aren't born on a Saturday is

6/7 * 6/7 . There are still 6 out of seven days possible

So three people not being born on a Saturday would be....

Answer 4

P(A union D)= 0.6, P(A)=0,4, if the question ask to compute P(D). the answer is 0,6-0.4 right??

Answer 5

P(AuD) = P(A) + P(D) - P(AnD), if P(AnD) = 0, then yes.

Answer 6

if they right P(AlB) what its mean??

Answer 7

Conditional probability.

It's the conditional probability that event A will occur given that event B has occurred

P(A|B)=P(AnB)/P(B)

or that event B will occur given that event A has occurred

P(B|A)=P(AnB)/P(A)

Answer 8

@tomatomoon

Conditional Probability

So imagine you're throwing two 6-sided dice,

you want condition A to be that both dice roll at least a 5, [5 5, 5 6, 6 5, 66]

P(A)=2/6 * 2/6 = 4/36 = 1/9

and you want condition B to be that the sum of the dice is odd
[1+2, 1+4, 1+6, 2+3, 2+5, 3+4, 3+6, 4+5, 5+6, 2+1, 4+1.....] 9*2 ways to do this

P(B) =18/36 = 1/2

P(AnB) is the probability of the intersection of A and B - the probability of conditions that satisfy both A and B. Here, there are only two. [5 6, 6 5], so

P(AnB) = 2/36 = 1/18

Now P(A|B) is the probability that A will occur given B, here it's saying "the probability that both dice are at least a 5, given that the sum of the rolls is odd." It satisfies both events, but in a particular order.

P(A|B) = P(AnB)/P(B) = (1/18)/(1/2) = 1/9

It's a 1 out of 9 chance that the dice will both be at least 5 given that the sum is odd (only two of the 18 outcomes satisfy A [5 6, 6 5].

P(B|A) = P(AnB)\P(A) = (1/18)/(1/9) = 1/2

The probability that the sum is odd given that both dice roll at least a 5 is 1/2. P(B|A) is much more likely than P(A|B) because there are only 4 possible outcomes given that A has occurred, wheres there are 18 possible outcomes given B has occurred.

:)

Answer 9

1. The Philadelphia office of Price Waterhouse Coopers hired five accounting trainees this
year. Their monthly starting salaries were: $3,536; $3,173; $3,448; $3,121; and $3,622.

2. A sample of 25 workers in a plant receive the hourly wages given in the table below.
3.65 3.78 3.85 3.95 4.00 4.10 4.25 3.55 3.85 3.96
3.60 3.90 4.26 3.75 3.95 4.05 4.08 4.15 3.80 4.05
3.88 3.95 4.06 4.18 4.05
question like this.. if they ask to fine variance. how can we know its use population variance or sample variance