Probability Tree (Dependent)

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Question 1 of 7

1. Question
A bowl has $8$ pieces of fruit, $5$ 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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Great Work!

Incorrect

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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 outcome

First 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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Correct

Well Done!

Incorrect

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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 outcome

First 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 colour

Write fractions in the format “a/b”
- $(a)$ (2/7, 12/42)
  
  $(b)$ (3/7, 18/42)
Hint

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Correct

Nice Job!

Incorrect

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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 outcome

First 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 $3/7$

$(a) 2/7$

$(b) 3/7$
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”
- $(a)$ (1/4, 3/12)
  
  $(b)$ (7/12)
Hint

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Correct

Excellent!

Incorrect

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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 ball from the box twice

Find the probabilities of each outcome

First 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 to 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 combined number divisible by $4$ is $1/4$

$(b)$    Find the probability of getting a number combination 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
$24$ is $7/12$

$(i) 1/4$

$(ii) 7/12$
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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Correct

Fantastic!

Incorrect

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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 outcome

First 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

Help Video

Correct

Keep Going!

Incorrect

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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 outcome

First 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)
  
  $(b)$ (2/3, 4/6)
  
  $(c)$ (1/3, 2/6)
Hint

Help Video

Correct

Correct!

Incorrect

Help Video

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 a three-student group.

Find the probabilities of each outcome

First 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 getting a number combination 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 drawing two discs with a combined number less than
$300$ is $2/3$

$(c)$    Find the probability of getting a number combination 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 drawing two discs with a combined number divisible
by $4$ is $1/3$

$(a) 1/6$

$(b) 2/3$

$(c) 1/3$

Probability Tree (Dependent)

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