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N76
2 months ago
12

A player is randomly dealt a sequence of 13 cards from a deck of 52-cards. All sequences of 13 cards are equally likely. In an e

quivalent model, the cards are chosen and dealt one at a time. When choosing a card, the dealer is equally likely to pick any of the cards that remain in the deck. What is the probability the 13th card dealt is a King?
Mathematics
1 answer:
zzz [12.3K]2 months ago
5 0

Response:

The answer is 1/13

To explain step-by-step:

We are notinformed about the first (01) twelve (12) cards that are shown; hence, the probability that the thirteenth (13) card dealt is a King is equivalent to the probability of the first card being a King, or specifically any designated card dealt being a King, equating to:

=4/52

=

1/13 FINAL RESULT

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Mandy is making stationery sets from 16 sheets of paper and 12 envelopes. If she wants all the sets to be identical without any
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4 sets Step-by-step explanation: The greatest common factor (GCF) of 16 and 12 is 4, which is the largest number that divides both evenly. This can be concluded from examining the factors: 12 = 4·3 and 16 = 4·4, or by confirming that the difference of the two numbers (4) can divide both without remainder. Hence, the maximum number of identical sets Mandy can create is 4. Each of these 4 sets will consist of 4 sheets of paper and 3 envelopes.
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2 months ago
Tags are placed to the left leg and right leg of a bear in a forest. Let A1 be the event that the left leg tag is lost and the e
babunello [11817]

Answer:

0.75 = 75% chance that only one tag is lost, provided at least one tag is lost

Step-by-step explanation:

Independent events:

If A and B are independent events, then:

P(A \cap B) = P(A)*P(B)

Conditional probability:

P(B|A) = \frac{P(A \cap B)}{P(A)}

Here

P(B|A) refers to the probability of event B occurring, given that event A has occurred.

P(A \cap B) is the probability of both A and B occurring together.

P(A) is the probability of event A occurring.

In this scenario:

Event A: At least one tag is missing

Event B: Only one tag is missing.

Each tag has a 40% likelihood of being lost, which is equal to 0.4.

Probability of at least one tag missing:

The events can be considered as either no tags are missing or at least one is. Their probabilities sum to 1. Thus

p + P(A) = 1

p is the probability that none are lost. Each tag has a 60% = 0.6 chance of not being lost, and since they are independent,

p = 0.6*0.6 = 0.36

Then

P(A) = 1 - p = 1 - 0.36 = 0.64

Intersection:

The intersection of at least one lost (A) and exactly one lost (B) is precisely one lost.

Then

Probability of at least one lost:

The first being lost (0.4 chance) and the second not lost (0.6 chance)

Or

The first not being lost (0.6 chance) and the second lost (0.4 chance)

So

P(A \cap B) = 0.4*0.6 + 0.6*0.4 = 0.48

Calculate the probability that exactly one tag is lost, given that at least one tag is lost (round to two decimal places).

P(B|A) = \frac{0.48}{0.64} = 0.75

0.75 = 75% likelihood that precisely one tag is lost, assuming at least one tag is lost

8 0
2 months ago
Hank and Debra each own two milking cows. One day, they milked their cows and compared the amount of milk the cows produced in t
AnnZ [12381]

Answer:

The result to your inquiry is 2 7/24.

Step-by-step explanation:

Data:

Hank's cows yield = 4 3/4 and 4 1/8

Debra's cows yield = 5 1/2 and 5 2/3

Process:

1.- Convert the mixed numbers into improper fractions:

For Hank: (16 + 3)/4 = 19/4 and (32 + 1)/8 = 33/8.

For Debra: (10 + 1)/2 = 11/2 and (15 + 2)/3 = 17/3.

2.- Add Hank's and Debra's milk quantities:

Hank: 19/4 + 33/8 = 38/8 + 33/8 = 71/8.

Debra: 11/2 + 17/3 = 33/6 + 34/6 = 67/6.

3.- Subtract Hank's total from Debra's total:

67/6 - 71/8 = (268 - 213)/24.

Thus, it simplifies to 55/24.

- Convert back to a mixed number:

2

24 55

7

Where 55/24 equals 2 7/24.

4.- Conclusion:

Debra produced 2 7/24 more milk than Hank.

4 0
2 months ago
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