Dependent Events
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Whenever you encounter probability questions involving two or more events happening at the same time or in sequence, you must first determine if their outcomes are independent or dependent.
Two events areÂ dependentÂ if the outcome of one event affects the probability of the other.
Here’s an example:
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For example, drawing a card from a deck and not returning it is an example of a dependent event. By not returning the card, you’ve decreased the total number of cards in the deck and the number of cards of that particular value (or suit, color, etc.) by one. This affects both the “bottom” and the “top” of probability calculations. If you draw a red seven, then there are 1 fewer sevens and 1 fewer red cards, so the probability of drawing a red or a seven is decreased.
Often the terms “not returned” or “without replacement” indicate that the events are dependent.
Dependent probabilities always coincide with “and” problems, so they will always be multiplication problems.
Solution
Since both cards must be green, this is an “and” question. Though the events are the same, the first card is not replaced, so they will have different probabilities. Calculate both probabilities then multiply.
First card
Bottom:Â 52 outcomes are possible
Top:Â 13 cards are green
Probability: \dfrac{13}{52} = \dfrac{1}{4}
Second card
Bottom:Â 51 outcomes are possible since the first card was not replaced
Top:Â 12 cards are green, since one green was drawn and not replaced
Probability: \dfrac{12}{51} = \dfrac{4}{17}
Probability of pulling 2 greens without replacement:
\dfrac{1}{4} Ã— \dfrac{4}{17} = \dfrac{1}{17}Solution
This is a dependent probability. Once the first marble is pulled, there is one less marble for the second pull.
First pull
Bottom:Â 10 marbles total
Top:Â 6 marbles are black
Probability: \dfrac{6}{10} = \dfrac{3}{5}
If the first marble was black, there will now only be 5 black marbles left, and 9 total marbles.
Second pull
Bottom:Â 9 marbles left
Top:Â 5 marbles are black
Probability: \dfrac{5}{9}
Total probability:
\dfrac{3}{5} Ã— \dfrac{5}{9} = \dfrac{3}{9} = \dfrac{1}{3}After each throw of a red die, the top face is marked with a blue stripe. What is the probability that after 6 throws all faces of the die will be marked blue?
A) \dfrac{3}{126}
B) \dfrac{5}{324}
C) \dfrac{1}{54}
D) \dfrac{2}{91}
E) \dfrac{5}{216}
Solution
Thinking through this problem logically will get you to the right answer. It is a dependent probability problem, because after each throw the number of sides the die can show decreases by 1. Remember to think bottom to top for each step.
First throw: None of the sides are marked, so the die can land any side up and it will work. The probability is \dfrac{6}{6} = 1. The top side gets marked. | |
Second throw:Â Only one side is marked, so the die can have any of the other 5 unmarked sides on top. The probability is \dfrac{5}{6}. The top gets marked. | |
Third throw:Â Two sides are marked, so the die can have any of the 4 unmarked sides on top. The probability is \dfrac{4}{6} = \dfrac{2}{3}. The top gets marked. | |
Fourth throw:Â Three sides are marked, so there are 3 unmarked sides. The probability is \dfrac{3}{6} = \dfrac{1}{2}. The top gets marked. | |
Fifth throw:Â Four sides are marked, so there are only 2 unmarked. The probability of is \dfrac{2}{6} = \dfrac{1}{3}. The top gets marked. | |
Sixth throw:Â Five sides are marked, so there is only one unmarked side. The probability is \dfrac{1}{6}. |
The probability is the product of the probabilities from each throw.
\dfrac{6}{6} Ã— \dfrac{5}{6} Ã— \dfrac{2}{3} Ã— \dfrac{1}{2} Ã— \dfrac{1}{3} Ã— \dfrac{1}{6}
= \dfrac{5}{6 Ã— 3 Ã— 3 Ã— 6}\\[3ex]
Simplify the fraction, butÂ wait before you calculate the denominator.
Notice that only two of the answers, (B) and (E), have 5 as the numerator. So either (B) or (E) is the answer.
Do a quick estimation of the denominator:
6 Ã— 3 Ã— 3 Ã— 6 â‰ˆ 6 Ã— 10 Ã— 6 = 360.
The denominator for (B) is 324 and for (E) is 216. Since 324 is closer to the estimate of 360, it is the best choice.
The correct answer is choice (B).
Dependent and Independent
You may encounter scenarios that combine both dependent and independent events. In this case you will multiply the probabilities of dependent outcomes and then add the probabilities of independent outcomes.
If a coin is tossed twice, what is the probability that it will land either both heads or both tails?
A) \dfrac{1}{16}
B) \dfrac{1}{8}
C) \dfrac{1}{4}
D) \dfrac{1}{2}
E) 1
Solution
This question has both “or” and “and” in the same problem: independent and dependent.
For the “or” part, there are two desired scenarios that could happen:Â H HÂ orÂ T T.
But within each scenario, there are two scenarios, and both must happen (bothÂ mustÂ be heads or bothÂ mustÂ be tails).
Both Heads:Â This means heads on the first toss and heads on the second toss. The probability of each toss coming up heads is
\dfrac{1}{2} Ã— \dfrac{1}{2} = \dfrac{1}{4}.
Both Tails:Â This has the same probability as heads:
\dfrac{1}{2} Ã— \dfrac{1}{2} = \dfrac{1}{4}
Since these are independent events, add the probabilities.
Both Heads or Both Tails:Â
\dfrac{1}{4} + \dfrac{1}{4} = \dfrac{2}{4} = \dfrac{1}{2}
The correct answer is choice (D).
One bag contains 4 blue and 5 red marbles. A second bag contains 3 blue and 4 red marbles. One marble is randomly pulled from the first bag and put into the second bag. After that, a marble is pulled from the second bag. What is the probability that this marble is blue?
A) \dfrac{1}{3}
B) \dfrac{5}{12}
C) \dfrac{31}{72}
D) \dfrac{4}{9}
E) \dfrac{11}{18}
Solution
There are just 2 possible scenarios.
1) A blue marble is pulled and put into the second bag and then a blue marble is pulled from the second bag.
2) A red marble is pulled and put into the second bag and then a blue marble is pulled from the second bag.
This question has both “or” and “and” in the same problem: independent and dependent.
For the “or” part, there are two desired scenarios that could happen: blue moved and blue pulled, or red moved and blue pulled.
But within each scenario, there are two scenarios, and both must happen (both blue moved and blue pulled, or both red moved and blue pulled).
Calculate the probability of each scenario and add them together.
The probability of the first scenario: The probability that a blue marble is pulled from the first basket is \dfrac{4}{9} (4 blue, 5 red). The probability that a blue marble is then pulled from the second basket is now \dfrac{4}{8} because there are now 4 blue marbles out of 8 total marbles. Multiply these dependent probabilities because it is an “and” scenario: \dfrac{4}{9} Ã— \dfrac{4}{8} = \dfrac{16}{72}. Leave this as an unsimplified fraction until addition.
The probability of the second scenario:Â The probability that a red marble is taken from the first bag is \dfrac{5}{9}. The probability that a blue marble is then taken from the second bag = \dfrac{5}{9} Ã— \dfrac{3}{8} = \dfrac{15}{72}.
The probability of EITHER the first OR the second scenarios:Â
\dfrac{16}{72} + \dfrac{15}{72} = \dfrac{31}{72}.
The correct answer is choice (C).
Section
Start by thinking about all the scenarios that could yield a sum of 7.
(1,6)Â Â Â (2,5)Â Â Â (3,4)
(6,1)Â Â Â (5,2)Â Â Â (4,3)
There are 6 pairs that sum to 7.
Method 1Â Â Â Â Using probabilities
Each pair has the probability of \dfrac{1}{6} Ã— \dfrac{1}{6} = \dfrac{1}{36}
These are 6 independent events, so “add” the probabilities by multiplying by 6.
6Â Ã— \dfrac{1}{36} = \dfrac{1}{6}
Method 2Â Â Â Â Using outcomes
There are a total of 36 possible pairs of dice. You could make a list of them all (see the end of this section), or use combinations (covered in the previous chapter) to discover this.
There are 6 pairs that sum to 7.
So the probability is:
\dfrac{desired \,outcomes}{total \,outcomes} = \dfrac{6}{36} = \dfrac{1}{6}Pairs of Dice
Pairs of dice are very common in probability questions. The table below shows the 36 possible pairs when rolling 2 dice that are the same color.
You can see that rolling a pair of dice and getting a 2 and a 4 can happen 2 ways: (2, 4) and (4, 2).
The probability of getting a 2 and a 4 is \dfrac{2}{36} = \dfrac{1}{18}.
(Red,Â Red) | (Red,Â Red) | (Red,Â Red) | (Red,Â Red) | (Red,Â Red) | (Red,Â Red) |
(1,Â 1) | (2,Â 1) | (3,Â 1) | (4,Â 1) | (5,Â 1) | (6,Â 1) |
(1,Â 2) | (2,Â 2) | (3,Â 2) | (4,Â 2) | (5,Â 2) | (6,Â 2) |
(1,Â 3) | (2,Â 3) | (3,Â 3) | (4,Â 3) | (5,Â 3) | (6,Â 3) |
(1,Â 4) | (2,Â 4) | (3,Â 4) | (4,Â 4) | (5,Â 4) | (6,Â 4) |
(1,Â 5) | (2,Â 5) | (3,Â 5) | (4,Â 5) | (5,Â 5) | (6,Â 5) |
(1,Â 6) | (2,Â 6) | (3,Â 6) | (4,Â 6) | (5,Â 6) | (6,Â 6) |
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