Dependent Events (Conditional Probability)

Question 1 of 7

A jar contains

7

$7$ orange and

6

$6$ green marbles. Find the probability of getting an orange, a green and another orange marble without replacement.

Write fractions in the format “a/b”

(21/143, 252/1716)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Probability Formula

$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$

Product Rule

$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$

Find the probability of getting an Orange marble.

favourable outcomes

$=$ $7$ (

$7$ Orange marbles)

total outcomes

$=$ $13$ (

$13$ total marbles)

$\mathsf{P(O_1)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{7}}{\color{#007DDC}{13}}$	Substitute values

Find the probability of getting a Green marble. Remember to reduce the total outcome.

favourable outcomes

$=$ $6$ (

$6$ Green marbles)

total outcomes

$=$ $12$ (

$1$ marble was already picked)

$\mathsf{P(G)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{6}}{\color{#007DDC}{12}}$	Substitute values

Find the probability of getting another Orange marble. Remember to reduce the number of Orange marbles and the total outcome.

favourable outcomes

$=$ $6$ (

$1$ Orange marble was already picked)

total outcomes

$=$ $6$ (

$2$ marbles were already picked)

$\mathsf{P(O_2)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{6}}{\color{#007DDC}{11}}$	Substitute values

Now, substitute the probabilities into the product rule

$\mathsf{P(O_1)}$	$=$	$\frac{7}{13}$

$\mathsf{P(G)}$	$=$	$\frac{6}{12}$

$\mathsf{P(O_2)}$	$=$	$\frac{6}{11}$

$\mathsf{P(Orange\:and\:Green\:and\:Orange)}$	$=$	$\mathsf{P(O_1)}\times\mathsf{P(G)}\times\mathsf{P(O_2)}$	Product Rule

	$=$	$\frac{7}{13}\times\frac{6}{12}\times\frac{6}{11}$	Substitute values

	$=$	$\frac{252}{1716}$

	$=$	$\frac{21}{143}$	Simplify

$21/143$

Question 2 of 7

Find the probability of drawing a Spade card then a Diamond card from a standard deck of cards without replacement.

Write fractions in the format “a/b”

(13/204, 169/2652)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Probability Formula

$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$

Product Rule

$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$

Find the probability of drawing a Spade card.

favourable outcomes

$=$ $13$ (

$13$ Spade cards)

total outcomes

$=$ $52$ (

$52$ total cards)

$\mathsf{P(Spade)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{13}}{\color{#007DDC}{52}}$	Substitute values

Find the probability of drawing a Diamond card. Remember to reduce the total outcome.

favourable outcomes

$=$ $13$ (

$13$ Diamond cards)

total outcomes

$=$ $51$ (

$1$ card was already drawn)

$\mathsf{P(Diamond)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{13}}{\color{#007DDC}{51}}$	Substitute values

Now, substitute the probabilities into the product rule

$\mathsf{P(Spade)}$	$=$	$\frac{13}{52}$

$\mathsf{P(Diamond)}$	$=$	$\frac{13}{51}$

$\mathsf{P(Spade\:and\:Diamond)}$	$=$	$\mathsf{P(Spade)}\times\mathsf{P(Diamond)}$	Product Rule

	$=$	$\frac{13}{52}\times\frac{13}{51}$	Substitute values

	$=$	$\frac{169}{2652}$

	$=$	$\frac{13}{204}$	Simplify

$13/204$

Question 3 of 7

Find the probability of drawing

$2$ Heart cards from a standard deck of cards without replacement.

Write fractions in the format “a/b”

(1/17, 156/2652)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Probability Formula

$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$

Product Rule

$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$

Find the probability of drawing a Heart card.

favourable outcomes

$=$ $13$ (

$13$ Heart cards)

total outcomes

$=$ $52$ (

$52$ total cards)

$\mathsf{P(H_1)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{13}}{\color{#007DDC}{52}}$	Substitute values

Find the probability of drawing another Heart card. Remember to reduce the number of Heart cards and the total outcome.

favourable outcomes

$=$ $12$ (

$1$ Heart card was already drawn)

total outcomes

$=$ $51$ (

$1$ card was already drawn)

$\mathsf{P(H_2)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{12}}{\color{#007DDC}{51}}$	Substitute values

Now, substitute the probabilities into the product rule

$\mathsf{P(H_1)}$	$=$	$\frac{13}{52}$

$\mathsf{P(H_2)}$	$=$	$\frac{12}{51}$

$\mathsf{P(Heart\:and\:Heart)}$	$=$	$\mathsf{P(H_1)}\times\mathsf{P(H_2)}$	Product Rule

	$=$	$\frac{13}{52}\times\frac{12}{51}$	Substitute values

	$=$	$\frac{156}{2652}$

	$=$	$\frac{1}{17}$	Simplify

$1/17$

Question 4 of 7

Find the probability of drawing a Jack and an Ace from a standard deck of cards without replacement.

Write fractions in the format “a/b”

(4/663, 16/2652)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Probability Formula

$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$

Product Rule

$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$

Find the probability of drawing a Jack card.

favourable outcomes

$=$ $4$ (

$4$ Jack cards)

total outcomes

$=$ $52$ (

$52$ total cards)

$\mathsf{P(Jack)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{4}}{\color{#007DDC}{52}}$	Substitute values

Find the probability of drawing an Ace card. Remember to reduce the total outcome.

favourable outcomes

$=$ $4$ (

$4$ Ace cards)

total outcomes

$=$ $51$ (

$1$ card was already drawn)

$\mathsf{P(Ace)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{4}}{\color{#007DDC}{51}}$	Substitute values

Now, substitute the probabilities into the product rule

$\mathsf{P(Jack)}$	$=$	$\frac{4}{52}$

$\mathsf{P(Ace)}$	$=$	$\frac{4}{51}$

$\mathsf{P(Jack\:and\:Ace)}$	$=$	$\mathsf{P(Jack)}\times\mathsf{P(Ace)}$	Product Rule

	$=$	$\frac{4}{52}\times\frac{4}{51}$	Substitute values

	$=$	$\frac{16}{2652}$

	$=$	$\frac{4}{663}$	Simplify

$4/663$

Question 5 of 7

Find the probability of drawing

$2$ Spade cards from a standard deck of cards without replacement.

Write fractions in the format “a/b”

(1/17, 156/2652)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Probability Formula

$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$

Product Rule

$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$

Find the probability of drawing a Spade card.

favourable outcomes

$=$ $13$ (

$13$ Spade cards)

total outcomes

$=$ $52$ (

$52$ total cards)

$\mathsf{P(S_1)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{13}}{\color{#007DDC}{52}}$	Substitute values

Find the probability of drawing another Spade card. Remember to reduce the number of Spade cards and the total outcome.

favourable outcomes

$=$ $12$ (

$1$ Spade card was already drawn)

total outcomes

$=$ $51$ (

$1$ card was already drawn)

$\mathsf{P(S_2)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{12}}{\color{#007DDC}{51}}$	Substitute values

Now, substitute the probabilities into the product rule

$\mathsf{P(S_1)}$	$=$	$\frac{13}{52}$

$\mathsf{P(S_2)}$	$=$	$\frac{12}{51}$

$\mathsf{P(Spade\:and\:Spade)}$	$=$	$\mathsf{P(S_1)}\times\mathsf{P(S_2)}$	Product Rule

	$=$	$\frac{13}{52}\times\frac{12}{51}$	Substitute values

	$=$	$\frac{156}{2652}$

	$=$	$\frac{1}{17}$	Simplify

$1/17$

Question 6 of 7

$50$ tickets were sold for a raffle and Jack bought

$10$ tickets. What is the probability of Jack winning the

$1st$ and

$2nd$ prize without replacement?

Write fractions in the format “a/b”

(9/245, 90/2450)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Probability Formula

$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$

Product Rule

$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$

Find the probability of Jack winning the

$1st$ prize.

favourable outcomes

$=$ $10$ (Jack’s tickets)

total outcomes

$=$ $50$ (Total tickets)

$\mathsf{P(1st)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{10}}{\color{#007DDC}{50}}$	Substitute values

Find the probability of Jack winning the

$2nd$ prize. Remember to reduce the number Jack’s tickets and the total outcome.

favourable outcomes

$=$ $9$ (

$1$ ticket was already used)

total outcomes

$=$ $49$ (

$1$ ticket was already used)

$\mathsf{P(2nd)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{9}}{\color{#007DDC}{49}}$	Substitute values

Now, substitute the probabilities into the product rule

$\mathsf{P(1st)}$	$=$	$\frac{10}{50}$

$\mathsf{P(2nd)}$	$=$	$\frac{9}{49}$

$\mathsf{P(First\:and\:Second)}$	$=$	$\mathsf{P(1st)}\times\mathsf{P(2nd)}$	Product Rule

	$=$	$\frac{10}{50}\times\frac{9}{49}$	Substitute values

	$=$	$\frac{90}{2450}$

	$=$	$\frac{9}{245}$	Simplify

$9/245$

Question 7 of 7

$50$ tickets were sold for a raffle and Jack bought

$10$ tickets. What is the probability of Jack not winning the

$1st$ and

$2nd$ prize without replacement?

Write fractions in the format “a/b”

(5/24, 10/48)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Probability Formula

$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$

Product Rule

$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$

Find the probability of Jack winning the

$1st$ and

$2nd$ prize.

favourable outcomes

$=$ $10$ (Jack’s tickets)

total outcomes

$=$ $48$ (

$1st$ and

$2nd$ prizes were removed)

$\mathsf{P(Not Winning)}$	$=$	$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$	Probability Formula

	$=$	$\frac{\color{#e65021}{10}}{\color{#007DDC}{48}}$	Substitute values

	$=$	$\frac{\color{#e65021}{5}}{\color{#007DDC}{24}}$	Simplify

Therefore, the probability of Jack not winning the

$1st$ and

$2nd$ prizes is

$5/24$

Dependent Events (Conditional Probability)

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

Excellent!

Probability Formula

Product Rule

2. Question

Hint

Well Done!

Probability Formula

Product Rule

3. Question

Hint

Nice Job!

Probability Formula

Product Rule

4. Question

Hint

Fantastic!

Probability Formula

Product Rule

5. Question

Hint

Correct!

Probability Formula

Product Rule

6. Question

Hint

Excellent!

Probability Formula

Product Rule

7. Question

Hint

Exceptional!

Probability Formula

Product Rule

Quizzes