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Mathematics 13 Online
OpenStudy (anonymous):

Probability help. What is the probability of drawing {a jack of spades, a 7 of clubs, a queen of hearts} from a standard deck of cards WITH replacement? A. 1/52 x 1/51 x 1/50 B. 1/8 C. (1/52)^3 D. 0.064

OpenStudy (anonymous):

replacement means you put back the card after you've drawn it right? so the probs don't change for 2nd, 3rd draw

OpenStudy (anonymous):

Right. So it can't be A, right?

OpenStudy (anonymous):

because I did not think about the probability of the cards I have no clue :D

OpenStudy (anonymous):

well jack spade is just once in there right ?

OpenStudy (anonymous):

every card is just once in there

OpenStudy (anonymous):

LMAO ohh lord. To be honest I have noooooo idea what a standard deck of cards is -.-

OpenStudy (anonymous):

So there's like one of each but different colours?

OpenStudy (anonymous):

Like black and red

OpenStudy (anonymous):

right I think every type of card is in all " 4colors" since they specify both then what they specify exists only once

OpenStudy (anonymous):

Well the color in the question is black I assume they mean only a black colour of each

OpenStudy (anonymous):

what's the # of cards in a deck? if we have this we can calculate how likely it is to draw exactly one card from it

OpenStudy (anonymous):

There are 52 cards in a standard deck according to Siri lol

OpenStudy (anonymous):

:D ok

OpenStudy (anonymous):

so since all of the cards they listed, I'm pretty sure only occur once in the deck then the chance to draw each of the cards is 1/52

OpenStudy (anonymous):

so when we draw the first card, we already have a 51/52 chance that we violated the correct sequence.

OpenStudy (anonymous):

and then there are 3 different card choices or whatever... does C make sense?

OpenStudy (anonymous):

and then when we actually managed to draw the correct card (just 1/52 chance!!) then we draw again and SURELY this time the 51/52 will stop us from completing the correct sequence

OpenStudy (anonymous):

we would need to get the 1/52 chance INSIDE OF THE INITIAL 1/52 chance, to make it to the last card

OpenStudy (anonymous):

only the best reach level 2, and then 51/52 of those that managed still die...

OpenStudy (anonymous):

\[\frac{ 1 }{ 52 } \times \frac{ 1 }{ 52 } \times \frac{ 1 }{ 52 }\]

OpenStudy (anonymous):

So ot's like practically impossible unless you a cheater/magician -.- lol I got like 0.0068 and a bunch of other numbers 0.o

OpenStudy (anonymous):

it's*

OpenStudy (anonymous):

yeah :) if you follow along a track of chances, you multiply them so that's why we multiplied each chance with each other, because we're only interested in the complete, correct trail. the trail is much less likely than getting one of the three.

OpenStudy (anonymous):

each probability has a 1/52 chance, but getting all the three 1/52 in the right order...... is equal to multiplying all three chances

OpenStudy (anonymous):

you can "simplify" this multiplication / write it without typing the same term thrice

OpenStudy (anonymous):

So to me it looks to be either C or D 0.o

OpenStudy (anonymous):

yep :)

OpenStudy (anonymous):

multiplying fractions works like this \[\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }= \frac{ 1 }{ 4 }\]

OpenStudy (anonymous):

(our probabilities ARE fractions, and they're multiplied so that's why it matters...)

OpenStudy (anonymous):

all following three are the same thing:\[\frac{ 1 }{ 2 } \times \frac{ 1 }{ 2 }=(\frac{ 1 }{ 2 })^2=\frac{ 1 }{ 2^2 }\]

OpenStudy (anonymous):

So it has to be in the form of a fraction?

OpenStudy (anonymous):

our probabilities are fractions

OpenStudy (anonymous):

well, you can transform fractions to the p= notation 1/4 = 25% = 0.25 in this case sticking to fractions makes most sense.

OpenStudy (anonymous):

ok what have we learned 1) 2) 3)

OpenStudy (anonymous):

So we're left with C! >:D

OpenStudy (anonymous):

1) there are 52 cards in a deck, and the card they want is one of a type chance to draw it is 1/52

OpenStudy (anonymous):

^right

OpenStudy (anonymous):

2) if you want a chain of probabilities you multiply them. we have a chain of three, so we multiply 1/52 three times in a row

OpenStudy (anonymous):

3) we can re-write 1/52*1/52*1/52 like 1/52^3 :)

OpenStudy (anonymous):

^check

OpenStudy (anonymous):

So it is C!!! >:D muahaha

OpenStudy (anonymous):

yep:) and in case you wondered if D was also possible- the decimal chance to get the cards in the right order is: 1 / 140608 or 7,11e-6 = 0.000007 or something

OpenStudy (anonymous):

only for magicians and cheaters.

OpenStudy (anonymous):

Ahhh ok. All is straight in my head for now. Thanks man much appreciation

OpenStudy (anonymous):

Lmfaooo!

OpenStudy (anonymous):

:)

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