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Mathematics 15 Online
OpenStudy (anonymous):

Marco draws a card and replaces it 10 times from a standard deck of 52 cards. He draws 8 red cards and 2 black cards. What is the theoretical probability that he will draw a red card on his 11th draw? 1/4 1/2 *** 2/3 4/5 @acxbox22 @david111

OpenStudy (campbell_st):

the theoretical probability of the next card being red is 1/2 and the previous results have no influence of the next draw... this is know as independent events

OpenStudy (anonymous):

@campbell_st thank you .

OpenStudy (anonymous):

A fire department has determined that on any particular day there is a 20% chance that it will have to respond to a call. What is the probability (to the nearest whole percent) that during any 3-day span it will have to respond to a call AT LEAST once?

OpenStudy (anonymous):

@david111 @acxbox22 @campbell_st

OpenStudy (campbell_st):

is is this a quiz you are doing...?

OpenStudy (anonymous):

@campbell_st . No it is a EOCT review sheet . ** hence the fact I am asking for HOW to go about solving these questions.

OpenStudy (anonymous):

the opposite of at least once is no call prob (no call) for 3 days = 0.8^3 prob ( at least one call) = 1 - 0.8^3 = 0.488 = 48.8 % answer B : 49 %

OpenStudy (campbell_st):

well use the complementary events P(At least 1) = 1 - P(0 calls for the 3 days)

OpenStudy (anonymous):

@david111 how did you get 0.8^3

OpenStudy (kirbykirby):

Probability of a call = 0.2 Probability of no call = 1 - 0.2 = 0.8 P(no call day 1) = 0.8 P(no call day 1 and 2) = P(no call day 1)*P(no call day 2) = 0.8*0.8 = 0.8^2 P(no call day 1 and 2 and 3 ) = 0.8*0.8*0.8 = 0.8^3

OpenStudy (anonymous):

@kirbykirby thank you for explaining that . so my answer is 49% ?

OpenStudy (acxbox22):

yes

OpenStudy (anonymous):

@david111 wait .. how did you know it was answer choice B ? I didn't post the choices .. lol

OpenStudy (kirbykirby):

Yes, but wonder where @david111 got option B from o_O?

OpenStudy (anonymous):

im a genius i can read ur mind

OpenStudy (anonymous):

@kirbykirby same here .. you copied and pasted lmaoo .

OpenStudy (anonymous):

nope i can see ur worksheet

OpenStudy (anonymous):

im in a tree with binoculars

OpenStudy (acxbox22):

stalk much

OpenStudy (acxbox22):

lol

OpenStudy (anonymous):

Mary Katherine draws a card and replaces it 50 times from a standard deck of 52 cards. She records her results and gets 20 red cards and 30 black cards. What is the experimental probability of selecting a black card on her next draw? 1 2/5 3/5 *** 3/4

OpenStudy (anonymous):

She's got 30 blacks and 20 reds. So her experimental probability is 30/50 = 0.6 (while the theoretical probability is 0.5).

OpenStudy (anonymous):

She got black 30/50 times so simplify to 3/5

OpenStudy (anonymous):

@david111 the answers are in fraction form .

OpenStudy (anonymous):

way ahead of you.

OpenStudy (anonymous):

choices?

OpenStudy (anonymous):

I already put them .

OpenStudy (acxbox22):

how many questions do you have left?

OpenStudy (anonymous):

3/5 is awsner.

OpenStudy (anonymous):

I have 9 left but I know how to do some of them on my own .

OpenStudy (anonymous):

responsibility

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