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Probability Tree (Dependent Events)Probability Tree (Dependent Events)
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Question 1 of 7
1. Question
A bowl has `8` pieces of fruit where `5` are bananas and `3` are oranges. Find the probability of drawing a banana and an orange in any order without replacement.Write fractions in the format “a/b”- (15/28, 30/56)
Hint
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$First, set up a probability tree showing all possible outcomes of drawing a fruit from the bowl twice
Find the probabilities of each outcomeFirst Stage – Drawing a Banana (B):`=` $$\frac{\color{#e65021}{5}}{\color{#007DDC}{8}}$$ `5` bananas out of `8` fruits Second Stage – Drawing a Banana (BB):`=` $$\frac{\color{#e65021}{4}}{\color{#007DDC}{7}}$$ `4` bananas out of `7` fruits Second Stage – Drawing an Orange (BO):`=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{7}}$$ `3` oranges out of `7` fruits First Stage – Drawing an Orange (O):`=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{8}}$$ `3` oranges out of `8` fruits Second Stage – Drawing a Banana (BB):`=` $$\frac{\color{#e65021}{4}}{\color{#007DDC}{7}}$$ `4` bananas out of `7` fruits Second Stage – Drawing an Orange (BO):`=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{7}}$$ `3` oranges out of `7` fruits Now, insert these probabilities to the probability tree
Pick the branches that lead to `1` Banana and `1` Orange then multiply the probabilities along them
First branch (BO):`5/8``times``3/7` `=` `15/56` Second branch (OB):`3/8``times``5/7` `=` `15/56` Finally, add the solved probability for each branch`15/56``+``15/56` `=` `30/56` `=` `15/28` Therefore, the probability of drawing a Banana and an Orange in any order is `15/28``15/28` -
Question 2 of 7
2. Question
A box contains $$4$$ cards, each with the letters $$\mathsf{W, X, Y, Z}$$.
A card is drawn from the box twice without replacement. Find the probability of drawing the letters $$\mathsf{W}$$ and $$\mathsf{Y}$$ in any order.Write fractions in the format “a/b”- (1/6, 2/12)
Hint
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$First, set up a probability tree showing all possible outcomes of drawing a card from the box twice
Find the probabilities of each outcomeFirst Stage – Drawing a W card:`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{4}}$$ `1` card out of `4` cards Second Stage – Drawing an X card (WX):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` card out of `3` cards Second Stage – Drawing a Y card (WY):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` card out of `3` cards Second Stage – Drawing a Z card (WZ):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` card out of `3` cards First Stage – Drawing an X card:`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{4}}$$ `1` card out of `4` cards Second Stage – Drawing a W card (XW):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` card out of `3` cards Second Stage – Drawing a Y card (XY):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` card out of `3` cards Second Stage – Drawing a Z card (XZ):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` card out of `3` cards First Stage – Drawing a Y card:`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{4}}$$ `1` card out of `4` cards Second Stage – Drawing a W card (YW):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` card out of `3` cards Second Stage – Drawing an X card (YX):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` card out of `3` cards Second Stage – Drawing a Z card (YZ):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` card out of `3` cards First Stage – Drawing a Z card:`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{4}}$$ `1` card out of `4` cards Second Stage – Drawing a W card (ZW):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` card out of `3` cards Second Stage – Drawing an X card (ZX):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` card out of `3` cards Second Stage – Drawing a Y card (ZY):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` card out of `3` cards Now, insert these probabilities to the probability tree
Pick the branches that lead to W and Y then multiply the probabilities along them
First branch (WY):`1/4``times``1/3` `=` `1/12` Second branch (YW):`1/4``times``1/3` `=` `1/12` Finally, add the solved probability for each branch`1/12``+``1/12` `=` `2/12` `=` `1/6` Therefore, the probability of drawing the cards with W and Y is `1/6``1/6` -
Question 3 of 7
3. Question
A box contains `7` balls. `4` are yellow and `3` are blue. Find the probability of drawing `2` balls without replacement and getting:`(a) 2` Yellow`(b)` Same colourWrite fractions in the format “a/b”-
`(a)` (2/7, 12/42)`(b)` (3/7, 18/42)
Hint
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$`(a)` Find the probability of getting `2` Yellow balls.First, set up a probability tree showing all possible outcomes of drawing a ball from the box twice
Find the probabilities of each outcomeFirst Stage – Drawing a Yellow ball:`=` $$\frac{\color{#e65021}{4}}{\color{#007DDC}{7}}$$ `4` Yellow balls out of `7` balls Second Stage – Drawing a Yellow (YY):`=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{6}}$$ `3` Yellow balls out of `6` balls Second Stage – Drawing a Blue ball (YB):`=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{6}}$$ `3` Blue balls out of `6` balls First Stage – Drawing a Blue ball:`=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{7}}$$ `3` Blue balls out of `7` balls Second Stage – Drawing a Yellow (BY):`=` $$\frac{\color{#e65021}{4}}{\color{#007DDC}{6}}$$ `4` Yellow balls out of `6` balls Second Stage – Drawing a Blue ball (BB):`=` $$\frac{\color{#e65021}{2}}{\color{#007DDC}{6}}$$ `3` Yellow balls out of `6` balls Now, insert these probabilities to the probability tree
Pick the branch that leads to `2` Yellow then multiply the probabilities along it
`4/7``times``3/6` `=` `12/42` `=` `2/7` Therefore, the probability of drawing `2` Yellow balls is `2/7``(b)` Find the probability of getting the same colour of balls.Use the probability tree from part `(a)` and pick the branches that lead to the same colour then multiply the probabilities along them
First branch (YY):`4/7``times``3/6` `=` `12/42` Second branch (BB):`3/7``times``2/6` `=` `6/42` Finally, add the solved probability for each branch`12/42``+``6/42` `=` `18/42` `=` `3/7` Therefore, the probability of drawing `2` balls of the same colour is `3/7``(a) 2/7``(b) 3/7` -
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Question 4 of 7
4. Question
A jar contains `4` discs labeled as `1, 2, 3` and `4`. Find the probability of drawing `2` discs without replacement and forming a two-digit number that is:`(a)` Divisible by `4``(b)` Greater than `23`Write fractions in the format “a/b”-
`(a)` (¼, 1/4, 3/12)`(b)` (7/12)
Hint
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Excellent!
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$`(a)` Find the probability of getting a number combination divisible by `4`.First, set up a probability tree showing all possible outcomes of drawing a disc from the jar twice
Find the probabilities of each outcomeFirst Stage – Drawing a `1`:`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{4}}$$ `1` disc out of `4` discs Second Stage – Drawing a `2` (`1 2`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` disc out of `3` discs Second Stage – Drawing a `3` (`1 3`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` disc out of `3` discs Second Stage – Drawing a `4` (`1 4`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` disc out of `3` discs First Stage – Drawing a `2`:`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{4}}$$ `1` disc out of `4` discs Second Stage – Drawing a `1` (`2 1`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` disc out of `3` discs Second Stage – Drawing a `3` (`2 3`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` disc out of `3` discs Second Stage – Drawing a `4` (`2 4`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` disc out of `3` discs First Stage – Drawing a `3`:`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{4}}$$ `1` disc out of `4` discs Second Stage – Drawing a `1` (`3 1`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` disc out of `3` discs Second Stage – Drawing a `2` (`3 2`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` disc out of `3` discs Second Stage – Drawing a `4` (`3 4`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` disc out of `3` discs First Stage – Drawing a `4`:`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{4}}$$ `1` disc out of `4` discs Second Stage – Drawing a `1` (`4 1`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` disc out of `3` discs Second Stage – Drawing a `2` (`4 2`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` disc out of `3` discs Second Stage – Drawing a `3` (`4 3`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` disc out of `3` discs Now, insert these probabilities into the probability tree
Pick the branches that lead to number combinations divisible by `4`, which are `12, 24,` and `32`
First branch (12):`1/4``times``1/3` `=` `1/12` Second branch (24):`1/4``times``1/3` `=` `1/12` Third branch (32):`1/4``times``1/3` `=` `1/12` Finally, add the solved probability for each branch`1/12``+``1/12``+``1/12` `=` `3/12` `=` `1/4` Therefore, the probability of drawing two discs with a number divisible by `4` is `1/4``(b)` Find the probability of forming a number greater than `23`.Use the probability tree from part `(a)` and pick the branches that lead to the number combinations greater than `23` then multiply the probabilities along them
First branch (24):`1/4``times``1/3` `=` `1/12` Second branch (31):`1/4``times``1/3` `=` `1/12` Third branch (32):`1/4``times``1/3` `=` `1/12` Fourth branch (34):`1/4``times``1/3` `=` `1/12` Fifth branch (41):`1/4``times``1/3` `=` `1/12` Sixth branch (42):`1/4``times``1/3` `=` `1/12` Seventh branch (43):`1/4``times``1/3` `=` `1/12` Finally, add the solved probability for each branch`1/12``+``1/12``+``1/12``+``1/12``+``1/12``+``1/12``+``1/12` `=` `7/12` Therefore, the probability of drawing two discs with a combined number greater than `23` is `7/12``(i) 1/4``(ii) 7/12` -
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Question 5 of 7
5. Question
A group of students consisting of `5` boys and `9` girls are forming a debating team. Find the probability of forming a team consisting of only `1` boy, without replacement.Write fractions in the format “a/b”- (45/91)
Hint
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Fantastic!
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$First, set up a probability tree showing all possible outcomes of forming a three-student group.
Find the probabilities of each outcomeFirst Stage – Assigning a Boy (B):`=` $$\frac{\color{#e65021}{5}}{\color{#007DDC}{14}}$$ `5` boys out of `14` students Second Stage – Assigning a Boy (BB):`=` $$\frac{\color{#e65021}{4}}{\color{#007DDC}{13}}$$ `4` boys out of `13` students Third Stage – Assigning a Boy (BBB):`=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{12}}$$ `3` boys out of `12` students Third Stage – Assigning a Girl (BBG):`=` $$\frac{\color{#e65021}{9}}{\color{#007DDC}{12}}$$ `9` girls out of `12` students Second Stage – Assigning a Girl (BG):`=` $$\frac{\color{#e65021}{9}}{\color{#007DDC}{13}}$$ `9` girls out of `13` students Third Stage – Assigning a Boy (BGB):`=` $$\frac{\color{#e65021}{4}}{\color{#007DDC}{12}}$$ `4` boys out of `12` students Third Stage – Assigning a Girl (BGG):`=` $$\frac{\color{#e65021}{8}}{\color{#007DDC}{12}}$$ `8` girls out of `12` students First Stage – Assigning a Girl (G):`=` $$\frac{\color{#e65021}{9}}{\color{#007DDC}{14}}$$ `9` girls out of `14` students Second Stage – Assigning a Boy (GB):`=` $$\frac{\color{#e65021}{5}}{\color{#007DDC}{13}}$$ `5` boys out of `13` students Third Stage – Assigning a Boy (GBB):`=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{12}}$$ `4` boys out of `12` students Third Stage – Assigning a Girl (GBG):`=` $$\frac{\color{#e65021}{8}}{\color{#007DDC}{12}}$$ `8` girls out of `12` students Second Stage – Assigning a Girl (GG):`=` $$\frac{\color{#e65021}{8}}{\color{#007DDC}{13}}$$ `8` girls out of `13` students Third Stage – Assigning a Boy (GGB):`=` $$\frac{\color{#e65021}{5}}{\color{#007DDC}{12}}$$ `5` boys out of `12` students Third Stage – Assigning a Girl (GGG):`=` $$\frac{\color{#e65021}{7}}{\color{#007DDC}{12}}$$ `7` girls out of `12` students Now, insert these probabilities to the probability tree
Pick the branches that lead to `1` Boy and `2` Girls then multiply the probabilities along them
First branch (BGG):`5/14``times``9/13``times``8/12` `=` `15/91` Second branch (GBG):`9/14``times``5/13``times``8/12` `=` `15/91` Third branch (GGB):`9/14``times``8/13``times``5/12` `=` `15/91` Finally, add the solved probability for each branch`15/91``+``15/91``+``15/91` `=` `45/91` Therefore, the probability of a group with only `1` Boy in any order is `45/91``45/91` -
Question 6 of 7
6. Question
A group of students consisting of `5` boys and `9` girls are forming a debating team. Find the probability of forming a team consisting of at least `1` girl, without replacement.Write fractions in the format “a/b”- (177/182)
Hint
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Keep Going!
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$First, set up a probability tree showing all possible outcomes of forming a three-student group.
Find the probabilities of each outcomeFirst Stage – Assigning a Boy (B):`=` $$\frac{\color{#e65021}{5}}{\color{#007DDC}{14}}$$ `5` boys out of `14` students Second Stage – Assigning a Boy (BB):`=` $$\frac{\color{#e65021}{4}}{\color{#007DDC}{13}}$$ `4` boys out of `13` students Third Stage – Assigning a Boy (BBB):`=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{12}}$$ `3` boys out of `12` students Third Stage – Assigning a Girl (BBG):`=` $$\frac{\color{#e65021}{9}}{\color{#007DDC}{12}}$$ `9` girls out of `12` students Second Stage – Assigning a Girl (BG):`=` $$\frac{\color{#e65021}{9}}{\color{#007DDC}{13}}$$ `9` girls out of `13` students Third Stage – Assigning a Boy (BGB):`=` $$\frac{\color{#e65021}{4}}{\color{#007DDC}{12}}$$ `4` boys out of `12` students Third Stage – Assigning a Girl (BGG):`=` $$\frac{\color{#e65021}{8}}{\color{#007DDC}{12}}$$ `8` girls out of `12` students First Stage – Assigning a Girl (G):`=` $$\frac{\color{#e65021}{9}}{\color{#007DDC}{14}}$$ `9` girls out of `14` students Second Stage – Assigning a Boy (GB):`=` $$\frac{\color{#e65021}{5}}{\color{#007DDC}{13}}$$ `5` boys out of `13` students Third Stage – Assigning a Boy (GBB):`=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{12}}$$ `4` boys out of `12` students Third Stage – Assigning a Girl (GBG):`=` $$\frac{\color{#e65021}{8}}{\color{#007DDC}{12}}$$ `8` girls out of `12` students Second Stage – Assigning a Girl (GG):`=` $$\frac{\color{#e65021}{8}}{\color{#007DDC}{13}}$$ `8` girls out of `13` students Third Stage – Assigning a Boy (GGB):`=` $$\frac{\color{#e65021}{5}}{\color{#007DDC}{12}}$$ `5` boys out of `12` students Third Stage – Assigning a Girl (GGG):`=` $$\frac{\color{#e65021}{7}}{\color{#007DDC}{12}}$$ `7` girls out of `12` students Now, insert these probabilities to the probability tree
Pick the branches that does not lead to at least `1` girl
First branch (BBB):`5/14``times``4/13``times``3/12` `=` `5/182` Finally, subtract the probability from `1``1-` `5/182` `=` `177/182` Therefore, the probability of a group with at least `1` Girl in any order is `177/182``177/182` -
Question 7 of 7
7. Question
A hat contains `3` cards labeled as `1, 2,` and `3`. Find the probability of drawing `3` cards without replacement and forming:`(a)` The number `231``(b)` A number less than `300``(c)` A number divisible by `4`Write fractions in the format “a/b”-
`(a)` (1/6)`(b)` (⅔, 2/3, 4/6)`(c)` (⅓, 1/3, 2/6)
Hint
Help VideoCorrect
Correct!
Incorrect
Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$`(a)` Find the probability of getting the number combination of `231`.First, set up a probability tree showing all possible outcomes of forming the number 231.
Find the probabilities of each outcomeFirst Stage – Drawing a `1` (`1`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` card out of `3` cards Second Stage – Drawing a `2` (`12`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` card out of `2` cards Third Stage – Drawing a `3` (`123`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{1}}$$ `1` card out of `1` card Second Stage – Drawing a `3` (`13`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` card out of `2` cards Third Stage – Drawing a `2` (`132`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{1}}$$ `1` card out of `1` card First Stage – Drawing a `2` (`2`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` card out of `3` cards Second Stage – Drawing a `1` (`21`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` card out of `2` cards Third Stage – Drawing a `3` (`213`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{1}}$$ `1` card out of `1` card Second Stage – Drawing a `3` (`23`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` card out of `2` cards Third Stage – Drawing a `1` (`231`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{1}}$$ `1` card out of `1` card First Stage – Drawing a `3` (`3`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ `1` card out of `3` cards Second Stage – Drawing a `1` (`31`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` card out of `2` cards Third Stage – Drawing a `2` (`312`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{1}}$$ `1` card out of `1` card Second Stage – Drawing a `2` (`32`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` card out of `2` cards Third Stage – Drawing a `1` (`321`):`=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{1}}$$ `1` card out of `1` card Now, insert these probabilities to the probability tree
There is only one branch that leads to the combination `231`. Multiply the probabilities along them.
`231` branch:`1/3``times``1/2``times``1/1` `=` `1/6` Therefore, the probability of a forming the number combination `231` is `1/6``(b)` Find the probability of forming a number less than `300`.Use the probability tree from part `(a)` and pick the branches that lead to the number combinations less than `300` then multiply the probabilities along them
First branch (123):`1/3``times``1/2``times``1/1` `=` `1/6` Second branch (132):`1/3``times``1/2``times``1/1` `=` `1/6` Third branch (213):`1/3``times``1/2``times``1/1` `=` `1/6` Fourth branch (231):`1/3``times``1/2``times``1/1` `=` `1/6` Finally, add the solved probability for each branch`1/6``+``1/6``+``1/6``+``1/6` `=` `4/6` `=` `2/3` Therefore, the probability of forming a number less than `300` is `2/3``(c)` Find the probability of forming a number divisible by `4`.Use the probability tree from part `(a)` and pick the branches that lead to the number combinations divisible by `4` then multiply the probabilities along them
First branch (132):`1/3``times``1/2``times``1/1` `=` `1/6` Second branch (312):`1/3``times``1/2``times``1/1` `=` `1/6` Finally, add the solved probability for each branch`1/6``+``1/6` `=` `2/6` `=` `1/3` Therefore, the probability of forming a number divisible by `4` is `1/3`.`(a) 1/6``(b) 2/3``(c) 1/3` -
Quizzes
- Simple Probability (Theoretical) 1
- Simple Probability (Theoretical) 2
- Simple Probability (Theoretical) 3
- Simple Probability (Theoretical) 4
- Complementary Probability
- Compound Events (Addition Rule) 1
- Compound Events (Addition Rule) 2
- Venn Diagrams (Mutually Inclusive)
- Independent Events 1
- Independent Events 2
- Dependent Events (Conditional Probability)
- Probability Tree (Independent Events) 1
- Probability Tree (Independent Events) 2
- Probability Tree (Dependent Events)