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3 years ago

Consider the example of finding the probability of selecting a black card or a 6 from a deck of 52 cards.

Mathematics

1 answer:

Marina86 [1]3 years ago

7 0

Answer:

The probability of selecting a black card or a 6 = 7/13

Step-by-step explanation:

In this question we have given two events. When two events can not occur at the same time,it is known as mutually exclusive event.

According to the question we need to find out the probability of black card or 6. So we can write it as:

P(black card or 6):

The probability of selecting a black card = 26/52

The probability of selecting a 6 = 4/52

And the probability of selecting both = 2/52.

So we will apply the formula of compound probability:

P(black card or 6)=P(black card)+P(6)-P(black card and 6)

Now substitute the values:

P(black card or 6)= 26/52+4/52-2/52

P(black card or 6)=26+4-2/52

P(black card or 6)=30-2/52

P(black card or 6)=28/52

P(black card or 6)=7/13.

Hence the probability of selecting a black card or a 6 = 7/13 ....

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Read 2 more answers

Four cards are dealt from a standard fifty-two-card poker deck. What is the probability that all four are aces given that at lea

elena-s [515]

Answer:

The probability is 0.0052

Step-by-step explanation:

Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:

P(A/B) = P(A∩B)/P(B)

The probability P(B) that at least three are aces is the sum of the following probabilities:

The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725

There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:

$nCk=\frac{n!}{k!(n-k)!}$

So, the number of ways to select exactly 3 aces is:

$4C3*48C1=\frac{4!}{3!(4-3)!}*\frac{48!}{1!(48-1)!}=192$

Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725

Then, the probability P(B) that at least three are aces is:

$P(B)=\frac{1}{270,725} +\frac{192}{270,725} =\frac{193}{270,725}$

On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:

P(A∩B) = 1/270,725

Finally, the probability P(A/B) that all four are aces given that at least three are aces is:

$P=\frac{1/270,725}{193/270,725} =\frac{1}{193}=0.0052$

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3 years ago