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Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$

= $\frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}$

= $\frac{6}{1081}$ = 0.0055

Now, probability that he ends up with a full house = $\frac{3}{1081} + \frac{6}{1081}$

= $\frac{9}{1081}$ = <u>0.0083</u>.

3 0

4 years ago

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I need help please I don’t know how to make this can someone help me ?

NISA [10]

Answer:

the simplified answer is 3 3/4

Step-by-step explanation:

8 0

3 years ago

Explain how you could estimate 22% of 78 to check if your answer is reasonable.

Sedbober [7]

Answer:

Step-by-step explanation:

you could think that 22% is close to 25% , so one forth of it, 1/4

78 percent is close to 80

1/4 of 80 is 20

actually 78*22/100=17.16

7 0

3 years ago

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Sunhee had four plastic shapes a square, a circle, and a pentagon in how many ways can she line up the four shapes of the circle

julsineya [31]

Answer:

only three shapes are mentioned.

But anyhow, 4 shapes allow the following reasoning...

you can start lining up with any of the for shapes, so you have 4 possibilities, for the second position you have 3 possibilities, for the third 2 and for the last only 1.

so 4*3*2*1= 24 possible ways

but with the added rule to keep the circle and the square apart, we have to reduce the number.

by the same step-by-step reasoning we get

4 possibilities for the 1st item

only 2 possibilities for the 2nd item

(if 1st is square or circle, the other one's not an option, if it's not we have to choose one of them to not place them both at point 3 and 4)

only 1 choice for the 3rd item

(reasoning as above, if 2nd is square or circle, we have to the only other, if it isn't, we can put it here at 3rd position)

plus scenarios where either square or circle are on 1st point, the other's on 4th.

I guess that makes 14 possible arrangements

5 0

3 years ago

Which of the following is the midsegment of abc<br><br> Help me please

Likurg_2 [28]

Answer:

Option C -LM

Step-by-step explanation:

Mid segment is a straight line joining the midpoints of two segments.A midsegment of a triangle is a segment connecting the midpoints of two sides of a triangle.In the given triangle ABC ,L and M are midpoints of sides AB and CB.LM is the line joining the midpoints of sides AB and CB.

LM is called the midsegment of triangle ABC.

5 0

3 years ago

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