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Find the probability of each event.Drawing two aces from a card deck without replacing the card after the first draw

$\frac{1}{169}$

Algebra

Chapter 8

Sequences, Series, and Probability

Section 7

Probability

Functions

Polynomials

Rational Functions

Introduction to Sequences and Series

Introduction to Combinatorics and Probability

Introduction to Algebra

Algebra and Trigonometry

Missouri State University

Oregon State University

McMaster University

Lectures

01:32

In mathematics, the absolu…

01:11

01:12

Find the probability of ea…

03:47

03:19

01:07

00:55

02:06

00:29

If two cards are drawn wit…

01:17

A card is drawn from a dec…

Two cards are drawn from a…

to figure out the probability that we will draw on a space on both hands without replacement. We're going to use the multiple could, that is, we're gonna find the probability of drawing an ace the first time and multiplied by the probability of Dragon. It's the second time during my first time I have 52 possible outcomes because I have 52 different cards. Four of these meet my requirements. That is their races. Now. I have now removed that one ace because I've taken it out on this firsthand. That leaves 51 possible outcomes for my 2nd 2nd draw on Lee, three of which meet my requirements. So I'm gonna be multiplying for 50 seconds times 3 51st Before we do this multiplication, though, let's simplify four. Divide by four is 1 52 Divided by four is 13. Likewise over here, three divided by three is one and 51. Divided by three is 70. When we multiply across the top, we get one and then we have 13 times 17 on the bottom 13 times 17 is 221 so you have a one out of 221 chance to drawn based on the first hand and then drawn ace on the second hand without replacement

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