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irakobra [83]
3 years ago
15

to determine the number of trout in a lake a lake a conservationist catches 124 trout tag them and throws them back into the lak

e later 44 trout are called 11 of them are tagged how many trout would the conservationist expect to be in the lake
Mathematics
1 answer:
natali 33 [55]3 years ago
8 0

Answer: 25

Step-by-step explanation:

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A large chess tournament starts with 1024 players. How many players will be remaining after 6 rounds
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16

Step-by-step explanation:

After reach round the number of players will be reduced by a half.

After 1 round it will be 1024 * 11/2 = 512, after  2 rounds it will be 1024 * (1/2)^2 = 256 etc..

Thus, after 6 rounds  the remaining players will be 1024* (1/2)^6

=  16  (answer)

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Please show work(15 points)
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We roll two fair 6-sided dice. Each of the 36 possible outcomes is assumed to be equally likely. (a) Given that the roll results
boyakko [2]

Answer:

a) \frac{1}{7}

b) \frac{2}{15}

Step-by-step explanation:

Given : We roll two fair 6-sided dice. Each of the 36 possible outcomes is assumed to be equally likely.

The outcomes are :

(1, 1) (1, 2)  (1, 3)  (1, 4)  (1, 5)  (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

(a) Given that the roll results in a sum of 3 or more, find the conditional probability that doubles (the first and the second rolls result in the same number) are rolled.

Let P(A) be the probability of event that the roll results in a sum of 3 or more.

Except (1,1) rest sum is 3 or greater than 3.

So, P(A)=\frac{35}{36}

Now, P(A and B) is that the roll results in a sum of 3 or more and that doubles (the first and the second rolls result in the same number) are rolled.

i.e. (2,2), (3,3), (4,4), (5,5), (6,6) - 5

P(A\cap B)=\frac{5}{36}

The conditional probability is given by,

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

P(B|A)=\frac{\frac{5}{36}}{\frac{35}{36}}

P(B|A)=\frac{5}{35}

P(B|A)=\frac{1}{7}

(b) Given that the two dice land on different numbers, find the conditional probability that the sum is 5.

Let P(A) be the probability of event that two dice land on different numbers.

Except (1,1),(2,2), (3,3), (4,4), (5,5), (6,6) rest two dice land on different numbers.

So, P(A)=\frac{30}{36}

Now, P(A and B) is that two dice land on different numbers and the sum is 5.

i.e. (1,4), (2,3), (3,2), (4,1) - 4

P(A\cap B)=\frac{4}{36}

The conditional probability is given by,

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

P(B|A)=\frac{\frac{4}{36}}{\frac{30}{36}}

P(B|A)=\frac{4}{30}

P(B|A)=\frac{2}{15}

4 0
3 years ago
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