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Question 1 of 7
1. Question
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)
Hint
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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
2. Question
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)
Hint
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Well Done!
Incorrect
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
3. Question
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)
Hint
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Nice Job!
Incorrect
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
4. Question
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)
Hint
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Fantastic!
Incorrect
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
5. Question
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)
Hint
Help VideoCorrect
Correct!
Incorrect
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
6. Question
`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)
Hint
Help VideoCorrect
Excellent!
Incorrect
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
7. Question
`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)
Hint
Help VideoCorrect
Exceptional!
Incorrect
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``5/24`