<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Basic Probability and Statistics Go to the latest version.

# 1.2: Dependent Events

Difficulty Level: At Grade Created by: CK-12

For 2 events to be dependent, the probability of the second event is dependent on the probability of the first event. In English, remember, the term dependent means to be unable to do without. This is similar to the mathematical definition of dependent events, where the second event is unable to happen without the first event occurring.

Example 6

Remember in Example 4 when you were asked to determine the probability of drawing 2 sevens from a standard deck of cards? What would happen if 1 card is chosen and not replaced? In this case, the probability of drawing a seven on the second draw is dependent on drawing a seven on the first draw. Now let’s calculate the probability of the 2 cards being drawn without replacement. This can be done with the Multiplication Rule.

Solution:

Let $A = 1^{\text{st}}$ seven chosen.

Let $B = 2^{\text{nd}}$ seven chosen.

$& 4 \ \text{suits} \qquad 1 \ \text{seven} \ \text{per suit}\\& \ \searrow \qquad \swarrow\\\text{The total number of sevens in the deck} &= 4 \times 1=4.$

$P(A) &= \frac{4}{52}\\\\& \qquad \qquad \ \ \text{Note: The total number of cards is}\\P(B) &= \frac{3}{51} \ \swarrow \text{51 after choosing the first card if}\\& \qquad \qquad \ \ \text{it is not replaced.}\\\\P(A \ \text{and} \ B) &= \frac{4}{52} \times \frac{3}{51} \ \text{or} \ P(A \cap B)=\frac{4}{52} \times \frac{3}{51}\\\\P(A \cap B) &= \frac{12}{2652}\\\\P(A \cap B) &= \frac{1}{221}$

Notice in this example that the numerator and denominator decreased from $P(A)$ to $P(B)$. Once we picked the first card, the number of cards available from the deck dropped from 52 to 51. The number of sevens also decreased from 4 to 3. Again, the explanation given in the example was that the first card chosen was kept in your hand and not replaced into the deck before the second card was chosen.

Example 7

A box contains 5 red marbles and 5 purple marbles. What is the probability of drawing 2 purple marbles and 1 red marble in succession without replacement?

Solution:

On the first draw, the probability of drawing a purple marble is:

On the third draw, the probability of drawing a red marble is:

$P(\text{red}) =\frac{5}{8}$

Therefore, the probability of drawing 2 purple marbles and 1 red marble is:

$P(1 \ \text{purple and 1 purple and 1 red}) &= P(1P \cap 1P \cap 1R)\\&= P_1(\text{purple}) \times P_2(\text{purple}) \times P(\text{red})\\ &= \frac{5}{10} \times \frac{4}{9} \times \frac{5}{8}\\&= \frac{100}{720}\\ &= \frac{5}{36}$

Feb 23, 2012

Jan 27, 2015

# Reviews

Image Detail
Sizes: Medium | Original

CK.MAT.ENG.SE.1.Prob-&-Stats-Basic-(Full-Course).1.2