+25 The probability of drawing two black cards without replacement is 23/94 , and the probability of drawing one black card is 1/2 . What is the probability of drawing a second black card, given that the first card is black?
23/47
2/3
1/3
23/94
My answer is 23/94 because it says in the question that drawing two black cars w/o replacement is 23/94 what do you guys think am I right? if not pleaes help step-by-step don't give me straight answers.
I'm going to go out on a limb here. I don't know formal methods of probability and statistics. But, I think as follows:
I'm assuming that we're using a standard deck of playing cards, without the jokers. 52 cards.
I understand why the probablilty of drawing 1 black card is 1/2. Half the cards are black, and the other half are red.
26/52 = 1/2 that's simple enough.
Now then, about drawing the next card.
What happened with the first card is independent of the second card. We aren't asking about two draws, only one draw.
After drawing a black card earlier, that leaves 25 black cards in the deck, and a total of 51 cards.
I say the chance of drawing a black card the second time is 25/51.
One of my teachers once told us that if you flip a coin and get 10 tails in a row, the odds of getting a tail of the 11th toss are still 1/2. What went before has no bearing on what comes next. That is not the same as asking the odds of flipping 11 tails in a row... which would be astronomical.
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+118609 The probability of drawing two black cards without replacement is 23/94 , and the probability of drawing one black card is 1/2 . What is the probability of drawing a second black card, given that the first card is black?
The prob of drawing a black card is a half so half the cards are black
Prob of drawing two balck cards = prob of drawing one * prob of drawing a second one = 23/94
so
\(\frac{1}{2}*P(second\; black) = \frac{23}{94}\\ P(second\; black) = \frac{23}{94}*2\\ P(second\; black) = \frac{23}{47}\\\)
