0.000990.00099incorrect 10.7861 0.786 , incorrect 0.999 incorrect At least one o

Question

0.000990.00099incorrect 10.7861 0.786 , incorrect 0.999 incorrect At least one of the answers above is NOT correct. (1 pt) A card is drawn from a standard deck of 52 cards, the value noted, and the card replaced. This is repeated 10 times. What is the probability that: a) Exactly 7 of the cards are diamonds? 0,00099 b) At least two of the cards are diamonds? 0.786 o) At most 7 of the cards are diamonds? 0.999 Note: You can earn partial credit on this problem. Preview s Submit Answers Your score was recorded. You have attempted this problem 8 times. Youreceived a score of 0% for this attempt. Your overall recorded score is 0% You have unlimited attempts remaining

Explanation / Answer

A)

Total Number of cards = 52

Number of diamond cards = 13

=> P(card drawn is diamond) = 1/4

All 10 draws are independent because the drawn card is replaced

=> P(Exactly 7 out of 10 are diamonds) = [(10C7)*(1/4)^7]*[(3/4)^3] = 0.0030899 [Answer A.]

Explanation

- 10C7 ways to choose 7 out of 10 draws (sequential draw) that yield a diamond card

- Probability of positive event occurring = 1/4 (7 times) and negative event occurring = 3/4 (3 times)

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B)

P(At least 2 are diamond) = 1 - P(1 or zero are diamond) = 1 - [P(Exactly 1 is diamond) + P(Exactly 0 are diamond)]

P(Exactly 1 is diamond) = (10C1)*[(1/4)^1]*[(3/4)^9] = 0.187712

Explanation

- 10C1 ways to choose 1 out of 10 draws (sequential draw) that yields a diamond card

- Probability of positive event occurring = 1/4 (1 times) and negative event occurring = 3/4 (9 times)

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P(Exactly 1 is diamond) = (10C0)*[(1/4)^0]*[(3/4)^10] = 0.0563135

Explanation

- 10C0 ways to choose 0 out of 10 draws (sequential draw) that yield a diamond card

- Probability of positive event occurring = 1/4 (0 times) and negative event occurring = 3/4 (10 times)

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=> P(At least 2 of the cards are diamonds) = 1 - ( 0.187712 + 0.0563135) = 0.755974 [Answer B.]

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C)

P(At most 7 are diamond) = 1 - P(8 or 9 or 10 are diamond) = 1 - [ P(Exactly 8 are diamond) + P(Exactly 9 are diamond) + P(Exactly 10 are diamond) ]

P(Exactly 8 are diamond) = (10C8)*[(1/4)^8]*[(3/4)^2] = 0.000386

Explanation

- 10C8 ways to choose 8 out of 10 draws (sequential draw) that yield a diamond card

- Probability of positive event occurring = 1/4 (8 times) and negative event occurring = 3/4 (2 times)

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P(Exactly 9 are diamond) = (10C9)*[(1/4)^9]*[(3/4)^1] = 0.0000286

Explanation

- 10C9 ways to choose 9 out of 10 draws (sequential draw) that yield a diamond card

- Probability of positive event occurring = 1/4 (9 times) and negative event occurring = 3/4 (1 time)

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P(Exactly 10 are diamond) = (10C10)*[(1/4)^10]*[(3/4)^0] = 0.00000095367

Explanation

- 10C10 ways to choose 10 out of 10 draws (sequential draw) that yield a diamond card

- Probability of positive event occurring = 1/4 (10 times) and negative event occurring = 3/4 (0 times)

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=> P(At most 7 of the cards are diamonds) = 1 - ( 0.000386 + 0.0000286 + 0.00000095367 )

= 0.9995844 [Answer C.]

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