CONDITIONAL PROBABILITY Assignment

Name: _________________________________________________ Period: ___________ Date: ________________
CONDITIONAL PROBABILITY Assignment
State which the following events are independent and which are dependent.
1. Drawing a card from a standard deck of playing card and flipping a penny
2. Drawing two disks from an jar without replacement of the first disk.
3. Winning the lottery and purchasing a new house
4. Driving on a high-speed highway and having an accident
5. Having a wide forehead and having a high IQ
A box contains 8 red coins, 9 yellow coins, and 5 blue coins. Two consecutive draws are made from the bag without
replacement of the first draw. Find the probability each of the following events.
6. Red first, red second
7. Red first, blue second
8. Red first, yellow second
9. Yellow first, blue second
10. Yellow first, yellow second
11. Yellow first, red second
12. Blue first, red second
13. Blue first, blue second
14. Blue first, yellow second.
For one roll of a die, let 𝑨 be the event β€œeven” and let 𝑩 the event β€œ4 or 6”. Find each probability.
15. 𝑷(𝑨)
16. 𝑷(𝑩)
17. 𝑷(𝑨 𝒂𝒏𝒅 𝑩)
18. 𝑷(𝑩 𝒂𝒏𝒅 𝑨)
19. 𝑷(𝑩|𝑨)
20. 𝑷(𝑨|𝑩)
𝟏
𝟏
21. Given 𝑷(𝑨 𝒂𝒏𝒅 𝑩) = and 𝑷(𝑨) = , find 𝑷(𝑩|𝑨).
πŸ’
𝟐
22. Given 𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝟎. πŸ‘πŸ– and 𝑷(𝑨) = 𝟎. πŸ“πŸ•, find 𝑷(𝑩|𝑨).
𝟏
𝟏
23. Given 𝑷(𝑩|𝑨) = πŸ‘ and 𝑷(𝑨) = 𝟐 , find 𝑷(𝑨 𝒂𝒏𝒅 𝑩).
Word Problem
24. A jar has 15 marbles, 3 of which are color red. If 2 are selected at random without replacing the first one, find
the probability that both are color red.
25. In a country club, the probability that a member who play tennis is 75%. If 54% of the members play tennis and
basketball, find the probability that the member who plays basketball, given that the member plays tennis.
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Name: _________________________________________________ Period: ___________ Date: ________________
CONDITIONAL PROBABILITY Assignment
ANSWER
State which the following events are independent and which are dependent.
1. Drawing a card from a standard deck of playing card and flipping a penny
Independent
2. Drawing two disks from an jar without replacement of the first disk.
Dependent
3. Winning the lottery and purchasing a new house
Dependent
4. Driving on a high-speed highway and having an accident
Dependent
5. Having a wide forehead and having a high IQ
Independent
A box contains 8 red coins, 9 yellow coins, and 5 blue coins. Two consecutive draws are made from the bag without
replacement of the first draw. Find the probability each of the following events.
6. Red first, red second
πŸ–
πŸ’
𝑷(𝑨) = 𝑷(𝒓𝒆𝒅) =
=
𝟐𝟐 𝟏𝟏
πŸ’ πŸ•
πŸ’ 𝟏
πŸ’
𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝑷(𝒓𝒆𝒅 𝒂𝒏𝒅 𝒓𝒆𝒅 ) =
β‹…
=
β‹… =
𝟏𝟏 𝟐𝟏 𝟏𝟏 πŸ‘ πŸ‘πŸ‘
πŸ’
𝑷(𝑨 𝒂𝒏𝒅 𝑩) πŸ‘πŸ‘
πŸ’ 𝟏𝟏 𝟏
𝑷(𝑩|𝑨) = 𝑷(𝒓𝒆𝒅 |𝒓𝒆𝒅 ) =
=
=
β‹…
=
πŸ’
𝑷(𝑨)
πŸ‘πŸ‘ πŸ’
πŸ‘
𝟏𝟏
𝑷(𝒓𝒆𝒅|𝒓𝒆𝒅 ) = πŸ‘πŸ‘. πŸ‘πŸ‘%
7. Red first, blue second
𝑷(𝑨) = 𝑷(𝒓𝒆𝒅) =
πŸ–
πŸ’
=
𝟐𝟐 𝟏𝟏
πŸ’ πŸ“
𝟐𝟎
β‹…
=
𝟏𝟏 𝟐𝟏 πŸπŸ‘πŸ
𝟐𝟎
𝑷(𝑨 𝒂𝒏𝒅 𝑩) πŸπŸ‘πŸ
𝟐𝟎 𝟏𝟏
πŸ“
𝑷(𝑩|𝑨) = 𝑷(𝒃𝒍𝒖𝒆 |𝒓𝒆𝒅 ) =
= πŸ’ =
β‹…
=
𝑷(𝑨)
πŸπŸ‘πŸ πŸ’
𝟐𝟏
𝟏𝟏
𝑷(𝒃𝒍𝒖𝒆|𝒓𝒆𝒅 ) = πŸπŸ‘. πŸ–πŸ%
𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝑷(𝒓𝒆𝒅 𝒂𝒏𝒅 𝒃𝒍𝒖𝒆 ) =
8. Red first, yellow second
𝑷(𝑨) = 𝑷(𝒓𝒆𝒅) =
πŸ–
πŸ’
=
𝟐𝟐 𝟏𝟏
πŸ’ πŸ—
πŸ’ πŸ‘ 𝟏𝟐
β‹…
=
β‹… =
𝟏𝟏 𝟐𝟏 𝟏𝟏 πŸ• πŸ•πŸ•
𝟏𝟐
𝑷(𝑨 𝒂𝒏𝒅 𝑩) πŸ•πŸ•
𝟏𝟐 𝟏𝟏 πŸ‘
𝑷(𝑩|𝑨) = 𝑷(π’šπ’†π’π’π’π’˜ |𝒓𝒆𝒅 ) =
= πŸ’ =
β‹…
=
𝑷(𝑨)
πŸ•πŸ• πŸ’
πŸ•
𝟏𝟏
𝑷(π’šπ’†π’π’π’π’˜|𝒓𝒆𝒅 ) = πŸ’πŸ. πŸ–πŸ”%
𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝑷(𝒓𝒆𝒅 𝒂𝒏𝒅 π’šπ’†π’π’π’π’˜ ) =
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CONDITIONAL PROBABILITY Assignment
9. Yellow first, blue second
𝑷(𝑨) = 𝑷(π’šπ’†π’π’π’π’˜) =
πŸ—
𝟐𝟐
πŸ— πŸ“
πŸπŸ“
β‹…
=
𝟐𝟐 𝟐𝟏 πŸπŸ“πŸ’
πŸπŸ“
𝑷(𝑨 𝒂𝒏𝒅 𝑩) πŸπŸ“πŸ’
πŸπŸ“ 𝟐𝟐
πŸ“
𝑷(𝑩|𝑨) = 𝑷(𝒓𝒆𝒅 |π’šπ’†π’π’π’π’˜) =
=
=
β‹…
=
πŸ—
𝑷(𝑨)
πŸπŸ“πŸ’ πŸ—
𝟐𝟏
𝟐𝟐
𝑷(𝒓𝒆𝒅|π’šπ’†π’π’π’π’˜) = πŸπŸ‘. πŸ–πŸ%
𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝑷(π’šπ’†π’π’π’π’˜π’‚π’π’… 𝒓𝒆𝒅 ) =
10. Yellow first, yellow second
𝑷(𝑨) = 𝑷(π’šπ’†π’π’π’π’˜) =
πŸ—
𝟐𝟐
πŸ— πŸ–
𝟏𝟐
β‹…
=
𝟐𝟐 𝟐𝟏 πŸ•πŸ•
𝟏𝟐
𝑷(𝑨 𝒂𝒏𝒅 𝑩) πŸ•πŸ•
𝟏𝟐 𝟐𝟐
πŸ–
𝑷(𝑩|𝑨) = 𝑷(𝒓𝒆𝒅 |π’šπ’†π’π’π’π’˜) =
= πŸ— =
β‹…
=
(
)
𝑷 𝑨
πŸ•πŸ• πŸ—
𝟐𝟏
𝟐𝟐
𝑷(𝒓𝒆𝒅|π’šπ’†π’π’π’π’˜) = πŸ‘πŸ–. πŸŽπŸ—%
𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝑷(𝒓𝒆𝒅 𝒂𝒏𝒅 π’šπ’†π’π’π’π’˜) =
11. Yellow first, red second
𝑷(𝑨) = 𝑷(π’šπ’†π’π’π’π’˜) =
πŸ—
𝟐𝟐
πŸ— πŸ–
𝟏𝟐
β‹…
=
𝟐𝟐 𝟐𝟏 πŸ•πŸ•
𝟏𝟐
𝑷(𝑨 𝒂𝒏𝒅 𝑩) πŸ•πŸ•
𝟏𝟐 𝟐𝟐
πŸ–
𝑷(𝑩|𝑨) = 𝑷(𝒓𝒆𝒅 |π’šπ’†π’π’π’π’˜) =
=
=
β‹…
=
πŸ—
𝑷(𝑨)
πŸ•πŸ• πŸ—
𝟐𝟏
𝟐𝟐
𝑷(𝒓𝒆𝒅|π’šπ’†π’π’π’π’˜) = πŸ‘πŸ–. πŸŽπŸ—%
𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝑷(π’šπ’†π’π’π’π’˜ 𝒂𝒏𝒅 𝒓𝒆𝒅 ) =
12. Blue first, red second
𝑷(𝑨) = 𝑷(𝒃𝒍𝒖𝒆 ) =
πŸ“
𝟐𝟐
πŸ“ πŸ–
𝟐𝟎
β‹…
=
𝟐𝟐 𝟐𝟏 πŸπŸ‘πŸ
𝟐𝟎
𝑷(𝑨 𝒂𝒏𝒅 𝑩) πŸπŸ‘πŸ
𝟐𝟎 𝟐𝟐
πŸ–
𝑷(𝑩|𝑨) = 𝑷(𝒓𝒆𝒅 |𝒃𝒍𝒖𝒆 ) =
=
=
β‹…
=
(
)
πŸ“
𝑷 𝑨
πŸπŸ‘πŸ πŸ“
𝟐𝟏
𝟐𝟐
𝑷(𝒓𝒆𝒅|𝒃𝒍𝒖𝒆 ) = πŸ‘πŸ–. πŸŽπŸ—%
𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝑷(𝒃𝒍𝒖𝒆 𝒂𝒏𝒅 𝒓𝒆𝒅 ) =
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CONDITIONAL PROBABILITY Assignment
13. Blue first, blue second
𝑷(𝑨) = 𝑷(𝒃𝒍𝒖𝒆 ) =
πŸ“
𝟐𝟐
πŸ“ πŸ’
𝟏𝟎
β‹…
=
𝟐𝟐 𝟐𝟏 πŸπŸ‘πŸ
𝟏𝟎
𝑷(𝑨 𝒂𝒏𝒅 𝑩) πŸπŸ‘πŸ
𝟏𝟎 𝟐𝟐
πŸ’
𝑷(𝑩|𝑨) = 𝑷(𝒃𝒍𝒖𝒆 |𝒃𝒍𝒖𝒆 ) =
=
=
β‹…
=
(
)
πŸ“
𝑷 𝑨
πŸπŸ‘πŸ πŸ“
𝟐𝟏
𝟐𝟐
𝑷(𝒃𝒍𝒖𝒆 |𝒃𝒍𝒖𝒆 ) = πŸπŸ—. πŸŽπŸ“%
𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝑷(𝒃𝒍𝒖𝒆 𝒂𝒏𝒅 𝒃𝒍𝒖𝒆 ) =
14. Blue first, yellow second.
𝑷(𝑨) = 𝑷(𝒃𝒍𝒖𝒆 ) =
πŸ“
𝟐𝟐
πŸ“ πŸ—
πŸ“ πŸ‘
πŸπŸ“
β‹…
=
β‹… =
𝟐𝟐 𝟐𝟏 𝟐𝟐 πŸ• πŸπŸ“πŸ’
πŸπŸ“
𝑷(𝑨 𝒂𝒏𝒅 𝑩) πŸπŸ“πŸ’
πŸπŸ“ 𝟐𝟐 πŸ‘
𝑷(𝑩|𝑨) = 𝑷(π’šπ’†π’π’π’π’˜|𝒃𝒍𝒖𝒆 ) =
=
=
β‹…
=
(
)
πŸ“
𝑷 𝑨
πŸπŸ“πŸ’ πŸ“
πŸ•
𝟐𝟐
𝑷(π’šπ’†π’π’π’π’˜|𝒃𝒍𝒖𝒆 ) = πŸ’πŸ. πŸ–πŸ”%
𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝑷(𝒃𝒍𝒖𝒆 𝒂𝒏𝒅 π’šπ’†π’π’π’π’˜) =
For one roll of a die, let 𝑨 be the event β€œeven” and let 𝑩 the event β€œ4 or 6”. Find each probability.
15. 𝑷(𝑨)
πŸ‘
𝟏
𝑷(𝑨) = = 𝑷(𝑨) =
πŸ”
𝟐
16. 𝑷(𝑩)
𝑷(𝑩) =
𝟏 𝟏 𝟐
𝟏
+ = = 𝑷(𝑩) =
πŸ” πŸ” πŸ”
πŸ‘
17. 𝑷(𝑨 𝒂𝒏𝒅 𝑩)
𝑷(𝑨 𝒂𝒏𝒅 𝑩) =
18. 𝑷(𝑩 𝒂𝒏𝒅 𝑨)
𝑷(𝑩 𝒂𝒏𝒅 𝑨) =
19. 𝑷(𝑩|𝑨)
20. 𝑷(𝑨|𝑩)
𝟏 𝟏
𝟏
β‹… = 𝑷(𝑨 𝒂𝒏𝒅 𝑩) =
𝟐 πŸ‘
πŸ”
𝟏 𝟏
𝟏
β‹… = 𝑷(𝑩 𝒂𝒏𝒅 𝑨) =
πŸ‘ 𝟐
πŸ”
𝟏
𝑷(𝑨 𝒂𝒏𝒅 𝑩) πŸ” 𝟏 𝟐
𝟏
𝑷(𝑩|𝑨 ) =
= 𝟏 = β‹… = 𝑷(𝑩|𝑨) =
𝑷(𝑨)
πŸ” 𝟏
πŸ‘
𝟐
𝟏
𝑷(𝑩 𝒂𝒏𝒅 𝑨) πŸ” 𝟏 πŸ‘
𝟏
𝑷(𝑨| 𝑩) =
= 𝟏 = β‹… = 𝑷(𝑨|𝑩) =
𝑷(𝑩)
πŸ” 𝟏
𝟐
πŸ‘
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CONDITIONAL PROBABILITY Assignment
𝟏
𝟏
21. Given 𝑷(𝑨 𝒂𝒏𝒅 𝑩) = and 𝑷(𝑨) = , find 𝑷(𝑩|𝑨).
πŸ’
𝟐
𝟏
𝑷(𝑨 𝒂𝒏𝒅 𝑩) πŸ’ 𝟏 𝟐
𝟏
𝑷(𝑩|𝑨) =
= 1 = β‹… = 𝑷(𝑩|𝑨) =
𝑷(𝑨)
πŸ’ 𝟏
𝟐
2
22. Given 𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝟎. πŸ‘πŸ– and 𝑷(𝑨) = 𝟎. πŸ“πŸ•, find 𝑷(𝑩|𝑨).
𝑷(𝑨 𝒂𝒏𝒅 𝑩) 𝟎. πŸ‘πŸ–
𝑷(𝑩|𝑨) =
=
= 𝟎. πŸ”πŸ”πŸ”πŸ” = 𝑷(𝑩|𝑨) β‰… πŸ”πŸ”. πŸ”πŸ•%
𝑷(𝑨)
𝟎. πŸ“πŸ•
𝟏
πŸ‘
𝟏
𝟐
23. Given 𝑷(𝑩|𝑨) = and 𝑷(𝑨) = , find 𝑷(𝑨 𝒂𝒏𝒅 𝑩).
𝟏 𝟏
𝟏
𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝑷(𝑩|𝑨) β‹… 𝑷(𝑨) = ( ) ( ) = = 𝑷(𝑨 𝒂𝒏𝒅 𝑩) β‰… πŸπŸ”. πŸ”πŸ”%
πŸ‘ 𝟐
πŸ”
Word Problem
24. A jar has 15 marbles, 3 of which are color red. If 2 are selected at random without replacing the first one, find
the probability that both are color red.
πŸ‘
𝟏
𝑷(𝑨) = 𝑷(𝒓𝒆𝒅) =
=
πŸπŸ“ πŸ“
𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝑷(𝒓𝒆𝒅 𝒂𝒏𝒅 𝒓𝒆𝒅 ) =
𝟏 𝟐
𝟏 𝟏
𝟏
β‹…
= β‹… =
πŸ“ πŸπŸ’ πŸ“ πŸ• πŸ‘πŸ“
𝟏
𝑷(𝑨 𝒂𝒏𝒅 𝑩) πŸ‘πŸ“
𝟏 πŸ“ 𝟏
𝑷(𝑩|𝑨) = 𝑷(𝒓𝒆𝒅|𝒓𝒆𝒅 ) =
= 𝟏 =
β‹… =
𝑷(𝑨)
πŸ‘πŸ“ 𝟏 πŸ•
πŸ“
𝑷(𝒓𝒆𝒅|𝒓𝒆𝒅 ) β‰… πŸπŸ’. πŸπŸ—%
25. In a country club, the probability that a member who play tennis is 75%. If 54% of the members play tennis and
basketball, find the probability that the member who plays basketball, given that the member plays tennis.
𝑷(𝑨) = 𝑷(π’•π’†π’π’π’Šπ’”) = 𝟎. πŸ•πŸ“
𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝑷(π’•π’†π’π’π’Šπ’” 𝒂𝒏𝒅 π’ƒπ’‚π’”π’Œπ’†π’•π’ƒπ’‚π’π’) = 𝟎. πŸ“πŸ’
𝑷(𝑩|𝑨) = 𝑷(π’ƒπ’‚π’”π’Œπ’†π’•π’ƒπ’‚π’π’|π’•π’†π’π’π’Šπ’” ) =
𝑷(𝑨 𝒂𝒏𝒅 𝑩) 𝟎. πŸ“πŸ’
=
= 𝟎. πŸ•πŸ
𝑷(𝑨)
𝟎. πŸ•πŸ“
𝑷(π’ƒπ’‚π’”π’Œπ’†π’•π’ƒπ’‚π’π’|π’•π’†π’π’π’Šπ’” ) = πŸ•πŸ%
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