Dependent Events (Conditional Probability)

Question 1 of 7

A jar contains

$7$ orange and

$6$ 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)

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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Quiz summary

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1. Question

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Excellent!

Probability Formula

Product Rule

2. Question

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Probability Formula

Product Rule

3. Question

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Nice Job!

Probability Formula

Product Rule

4. Question

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Fantastic!

Probability Formula

Product Rule

5. Question

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Correct!

Probability Formula

Product Rule

6. Question

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Excellent!

Probability Formula

Product Rule

7. Question

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Exceptional!

Probability Formula

Product Rule

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