Find the probability of drawing 3 Aces at random from a deck of 52 ordinary playing cards if the cards are:_________

Mathematics

1 answer:

Anuta_ua [19.1K]3 years ago

7 0

Answer:

a. With replacement

1/2197

b. Without replacement

1/5,525

Step-by-step explanation:

Okay, here is a probability question.

The key to answering this question is by knowing the number of aces in a deck of cards.

There is 1 ace per suit, so there is a total of 4 aces per deck of cards.

So, mathematically the probability of picking an ace would be;

number of aces/ total number of cards = 4/52 = 1/13

a. Now since the action is with replacement; that means that at any point in time, the total number of cards would always remain 52 even after making our picks.

So the probability of picking three aces with replacement would be;

1/13 * 1/13 * 1/13 = 1/2197

b. Without replacement

what this action means is that after picking a particular card, we do not return the picked card to the deck of cards.

For the first card picked, we will be having a total of 4 aces and 52 total cards.

So the probability of picking an ace would be 4/52 = 1/13

For the second card picked, we shall be left with selecting an ace out of the remaining 3 aces and the total remaining 51 cards

So the probability will be 3/51 = 1/17

For the third and last card to be picked, we shall be left with picking 1 out of the remaining 2 aces cards and out of the 50 cards left in the deck.

So the probability now becomes 2/50 = 1/25

Thus, the combined probability of picking 3 aces cards without replacement from a deck of cards will be;

1/13 * 1/17 * 1/25 = 1/5,525

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Answer:

$\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}$

Step-by-step explanation:

$\left(2^8\cdot5^{-5}\cdot19^0\right)^{-2}\cdot\left(\dfrac{5^{-2}}{2^3}\right)^4\cdot228\\\\\text{use}\ a^{-n}=\dfrac{1}{a^n}\ \text{and}\ a^0=1\ \text{and}\ (a^n)^m=a^{nm}\\\\=\left(2^8\cdot\dfrac{1}{5^5}\cdot1\right)^{-2}\cdot\left(\dfrac{\frac{1}{5^2}}{2^3}\right)^4\cdot228=\left(\dfrac{2^8}{5^5}\right)^{-2}\cdot\left(\dfrac{1}{2^35^2}\right)^4\cdot228$

$=\dfrac{(2^8)^{-2}}{(5^5)^{-2}}\cdot\dfrac{1^4}{(2^3)^4(5^2)^4}\cdot228=\dfrac{2^{-16}}{5^{-10}}\cdot\dfrac{1}{2^{12}5^8}\cdot228\\\\\text{use}\ a^n=\dfrac{1}{a^{-n}}\to\dfrac{1}{a^n}=a^{-n}\\\\=2^{-16}\cdot5^{10}\cdot2^{-12}\cdot5^{-8}\cdot228\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=2^{-16+(-12)}\cdot5^{10+(-8)}\cdot228=2^{-28}\cdot5^2\cdot228\\\\=2^{-28}\cdot5^2\cdot4\cdot57=2^{-28}\cdot5^2\cdot2^2\cdot57=2^{-28+2}\cdot5^2\cdot57\\\\=2^{-26}\cdot5^2\cdot57=\dfrac{5^2\cdot57}{2^{26}}$