Question

If you draw a card at random from a deck of playing cards what is the probability it is a heart, spade, club or diamond?

Answer 1

@jim_thompson5910 can u help me?

Answer 2

The probability is 1 or 100% because there are only four suits and they are hearts, spades, clubs or diamonds.

So it's guaranteed you're going to draw one of the four options.

Answer 3

okay and can u help me out with a couple more. I need help plese,and your good at explaining.

Answer 4

sure whats up

Answer 5

If you dial a seven digit number at random on a phone, what is the probability you dial your next door neighbor's main phone? for this is the answer 1/5040

Answer 6

Assuming all digits are likely to be dialed at any one time, there are 10^7 = 10,000,000 possible phone numbers to dial


But the neighbor only has one phone number (assuming s/he has only one land-line or cell). So the probability is 1/10000000

Answer 7

Your answer is assuming that you can't re-use any digits but numbers like 555-1212 are perfectly possible.

Answer 8

oh okay and for thjis question : A shootout in hockey consists of Teams A and B taking alternate shots on goal. Suppose that the rules state that the first team to score wins. Team A has a probability of 0.39 of scoring with any one shot. Team B has a probability of 0.39 of scoring with any one shot. If team A shoots first, what is the probability, to 2 decimal places, of Team B winning on its first shot? is the answer 0.24

Answer 9

Team A shoots first

So the probability of team A winning on the first shot is given as 0.39

In order for B to win on the first shot, A must NOT win on the first shot (otherwise, the game is over)


So the probability of A NOT winning on the first shot is 1 - 0.39 = 0.61


This then means that...


P(B winning on first shot) = P(A NOT winning on first shot AND B winning on first shot)


P(B winning on first shot) = P(A NOT winning on first shot)*P(B winning on first shot)


P(B winning on first shot) = 0.61*0.39


P(B winning on first shot) = 0.2379


P(B winning on first shot) = 0.24


So you are correct

Answer 10

okay and what about this question: City Council consists of 8 men and 5 women. If 6 representatives are chosen at random to form an environmental sub-committee, what is the probability that Mayor Mr. Bill Tompson and 3 women are chosen, correct to the nearest thousandth?

Answer 11

There are 8 men + 5 women = 13 people total


So there are 13 C 6 = 1716 different ways to form a sub-committee (with no restrictions)


Since the mayor must be on the sub-committee, there are now 6-1 = 5 places left to fill. There are also 13 - 1 = 12 people left to choose from.


There are 5 women, and 3 women must be on the committee, so there are 5 C 3 = 10 different ways to add on 3 women.


The remaining 2 spots left and these spots must be filled by men (since there must be exactly 3 women). So there are 7 C 2 = 21 ways to do this


So overall there are 1*10*21 = 210 different ways to form the committee that has exactly 3 women and the mayor (and 2 other guys)


So the probability is 210/1716 = 35/286 = 0.1223776

Answer 12

From a group of 13 student representatives, 5 will be chosen to work on a committee. What is the probability that Sally is chosen, correct to the nearest thousandth? is the answer 1/1287

Answer 13

so the answer to the nearest thousandth would be 0.001

Answer 14

There are 13 C 5 = 1287 different ways to form this committee


Sally must be chosen, so there are 5-1 = 4 slots left. There are 13-1 = 12 students left to choose from. So there are 12 C 4 = 495 ways to choose a committee where Sally is on it.


So the probability is 495/1287 = 5/13 = 0.3846

Answer 15

oh okay i realized my mistake. and 2 last question if you dont mind: At a family reunion, door prizes are to be given out. At one table in the community hall, 6 children, 3 teenagers, 5 adults, and 6 seniors are seated. The 3 winning tickets are held by 3 different people at this table. is the answer for this 0.045

Answer 16

oh never mind i got it thanks for the help, ill probably be asking u more questions if your sticking around. lol

Answer 17

It'll depend really, so try to get as much done and ask the ones that are really bugging you.

Answer 18

hey can u help?

Answer 19

A basketball player has scored on 60% of her free throws throughout her career. Given this information, the probability that she will score on exactly 9 of her next 14 free throws, correct to the nearest hundredth, is

Answer 20

Use the binomial distribution probability formula

Answer 21

P(X = x) = (n C x)*(p)^(x)*(1-p)^(n-x)

P(X = 9) = (14 C 9)*(0.6)^(9)*(1-0.6)^(14-9)

P(X = 9) = (14 C 9)*(0.6)^(9)*(0.4)^(14-9)

P(X = 9) = (2002)*(0.6)^(9)*(0.4)^5

P(X = 9) = (2002)*(0.010077696)*(0.01024)

P(X = 9) = 0.20659760529408

So the probability that she will score on exactly 9 of her next 14 free throws is roughly 0.20659760529408

Answer 22

okay thts wht i gowht i got too how abt this one: An examination consists of 50 multiple choice questions each with four possible answers. What is the probability of guessing exactly 10 correct answers?

Answer 23

Again use the binomial distribution probability formula


P(X = x) = (n C x)*(p)^(x)*(1-p)^(n-x)

P(X = 10) = (50 C 10)*(0.25)^(10)*(1-0.25)^(50-10)

P(X = 10) = (50 C 10)*(0.25)^(10)*(0.75)^(50-10)

P(X = 10) = (10272278170)*(0.25)^(10)*(0.75)^40

P(X = 10) = (10272278170)*(0.00000095367)*(0.000010056585)

P(X = 10) = 0.09851796245376


So the probability of guessing exactly 10 correct answers is 0.09851796245376

Answer 24

am i bugging u buy asking all these questions,cause i can ask someone else? cause i have 3 more questions to ask.

Answer 25

you can ask me, i'm just going to be slow to respond because I'm answering other questions

Answer 26

so it's up to you

Answer 27

okay ill ask u then: A binomial experiment consists of 6 trials with a probability of success on each trial of 0.2. Calculate the probability of obtaining at least 3 successes.

Answer 28

is the answer 0.099

Answer 29

Approximately 3% of peanuts in a certain shipment are spoiled. If you randomly select 24 peanuts, what is the probability that at least one peanut is spoiled? and this one

Answer 30

one question at a time please

Answer 31

it's hard to reference multiple questions

Answer 32

oh sorry about that.

Answer 33

no worries, this is the answer to the first one


P(X >= 3) = P(X = 4) + P(X = 5) + P(X = 6)


So we need the following P(X = 4), P(X = 5), and P(X = 6)


P(X = x) = (n C x)*(p)^(x)*(1-p)^(n-x)

P(X = 4) = (6 C 4)*(0.2)^(4)*(1-0.2)^(6-4)

P(X = 4) = (6 C 4)*(0.2)^(4)*(0.8)^(6-4)

P(X = 4) = (15)*(0.2)^(4)*(0.8)^2

P(X = 4) = (15)*(0.0016)*(0.64)

P(X = 4) = 0.01536


P(X = x) = (n C x)*(p)^(x)*(1-p)^(n-x)

P(X = 5) = (6 C 5)*(0.2)^(5)*(1-0.2)^(6-5)

P(X = 5) = (6 C 5)*(0.2)^(5)*(0.8)^(6-5)

P(X = 5) = (6)*(0.2)^(5)*(0.8)^1

P(X = 5) = (6)*(0.00032)*(0.8)

P(X = 5) = 0.001536


P(X = x) = (n C x)*(p)^(x)*(1-p)^(n-x)

P(X = 6) = (6 C 6)*(0.2)^(6)*(1-0.2)^(6-6)

P(X = 6) = (6 C 6)*(0.2)^(6)*(0.8)^(6-6)

P(X = 6) = (1)*(0.2)^(6)*(0.8)^0

P(X = 6) = (1)*(0.000064)*(1)

P(X = 6) = 0.000064


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P(X >= 3) = P(X = 4) + P(X = 5) + P(X = 6)


P(X >= 3) = 0.01536 + 0.001536 + 0.000064


P(X >= 3) = 0.01696


So the probability of getting at least 3 successes is 0.01696

Answer 34

Well I guess I could do "Question: ...., then Answer:...", but it's still a good idea to do this one at a time.


Question:

Approximately 3% of peanuts in a certain shipment are spoiled. If you randomly select 24 peanuts, what is the probability that at least one peanut is spoiled?

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Answer:

P(At least one spoiled) = 1 - P(None spoiled)

P(At least one spoiled) = 1 - (1-0.03)^24

P(At least one spoiled) = 1 - (0.97)^24

P(At least one spoiled) = 1 - 0.48141722191172

P(At least one spoiled) = 0.51858277808828

Answer 35

okay thankyou so much jim, i really appreciate it!

Answer 36

you're welcome