Two cards are drawn in succession without replacement from a standard deck of 52 cards. What is the probability that the first card is a spade given that the second card is a diamond? (Round your answer to three decimal places.)
Again, go through the same processes
Let
S=event that the first card drawn is a spade
~S=event that the first card drawn is not a spade
s=event that the second card drawn is a spade
~s=event that the second card drawn is not a spade
then
probability of NOT drawing a spade in the second draw, given the first card drawn was a spade
P(~s|S)=39/51
probability of not drawing a spade in the second draw, given the first card drawn was NOT a spade
P(~s|~S)=38/51
We already know that the probability of drawing a space in the first card is
P(S)=13/52=1/4
Total probability of NOT drawing a spade in the second draw (any of diamond, club, heart)
P(~s)=P(~s|S)P(S)+P(~s|~S)P(~S)
=39/51*(1/4)+38/51*(3/4)
=3/4
Using Bayes theorem
P(S|~s)=P(~s|S)*P(S)/P(~s)
=(39/51)*(1/4)/(3/4)
=13/51
(slightly better than the intuitive value of 13/52=1/4).
WHY WOULD THE PROBABILITY OF THE FIRST CARD DRAWN BE AFFECTED BY THE POSTERIOR INFORMATION THAT THE SECOND CARD IS NOT A SPADE?
Imagine the extreme case where we know the following 39 draws are NOT spades, then we know that the first card drawn MUST be a spade, with probability 1.
Therefore posterior information can affect the probability of the first card drawn.
Well, let's think about this statistically, but with a touch of humor!
If we're picking two cards from a standard deck, without replacement, we know that there are 52 total cards to begin with.
Now, suppose we've already drawn the second card and, lo and behold, it's a diamond! How dazzling!
Before this, there were 13 diamonds in the deck. So, the probability of drawing any random card and it being a diamond was 13/52 or 1/4. Now, since we've already established that the second card is a diamond, we can safely assume that there are only 12 diamonds left in the deck for the first card.
Were you paying attention? Good! There are 12 spades in total, one of which will be the first card we draw. So, the probability that the first card is a spade, given that the second card is a diamond, is 12/51, because now there are only 51 cards left in the deck.
Let's put some numbers to my clowning around. The probability is approximately 0.235 when rounded to three decimal places.
So, the answer is 0.235, or in clown terms, it's like saying there's about a 23.5% chance the first card is a spade after you've struck gold with a diamond on the second round!
Now, go spread some laughter while dealing cards!
To find the probability that the first card is a spade given that the second card is a diamond, we can use the concept of conditional probability.
Let's break down the problem step by step:
Step 1: Determine the total number of ways we can draw two cards from a standard deck of 52 cards without replacement. This is denoted as C(52, 2), which is the combination of 52 choose 2:
C(52, 2) = (52!)/((2!)(52-2)!) = 1326
Step 2: Determine the number of ways we can draw a diamond as the second card. There are 13 diamonds in a deck of 52 cards.
Step 3: Determine the number of ways we can draw a spade as the first card and a diamond as the second card. There are 13 spades in a deck of 52 cards.
Step 4: Apply the formula for conditional probability:
P(spade|diamond) = P(spade and diamond) / P(diamond)
Step 5: Calculate the probability:
P(spade and diamond) = (Number of ways to draw a spade and a diamond) / (Number of ways to draw two cards without replacement)
P(spade and diamond) = (13/52) * (13/51) [Since we have 13 ways to draw a spade and 13 ways to draw a diamond as the second card]
P(diamond) = (Number of ways to draw a diamond as the second card) / (Number of ways to draw two cards without replacement)
P(diamond) = (13/52) * (12/51) [Since we have 13 ways to draw a diamond as the second card and 12 ways to draw any card for the first card]
P(spade|diamond) = P(spade and diamond) / P(diamond)
P(spade|diamond) = ((13/52) * (13/51)) / ((13/52) * (12/51))
Simplifying this expression:
P(spade|diamond) = (13/51) / (12/51)
P(spade|diamond) = 13/12
Finally, rounding the answer to three decimal places:
P(spade|diamond) ≈ 1.083
Therefore, the probability that the first card is a spade given that the second card is a diamond is approximately 1.083.
To find the probability that the first card is a spade given that the second card is a diamond, we can make use of conditional probability.
Let's break down the problem step by step:
1. Determine the total number of cards in the deck to start with, which is 52.
2. The probability of drawing a diamond card on the second draw, given that the first card is a spade, can be obtained by reducing the size of the deck by one (since we are drawing the second card without replacement). The total number of cards remaining will be 51.
3. To calculate the probability, we divide the number of favorable outcomes (spades) by the number of possible outcomes (remaining cards in the deck).
4. There are 13 spade cards in a deck of 52 cards, so the probability of drawing a spade first is 13/52.
5. After removing the first card, the number of diamond cards remaining in the deck is 13 since there are 13 diamonds in a standard deck.
6. Plugging these values into the conditional probability formula: P(spade first | diamond second) = (P(spade first) * P(diamond second)) / P(diamond second) = (13/52 * 13/51) / (13/51) = 13/52.
After simplifying the fraction, we find that the probability that the first card is a spade given that the second card is a diamond is 13/52, which is equivalent to 0.250 when rounded to three decimal places.