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Blizzard [7]
3 years ago
6

A fair coin and a biased coin where the probability of obtaining a head is twice the probability of obtaining a tail. if both ar

e tossed simultaneously, find the probability that they are:
a) both heads
b) both tails
c) one head
d) at least one head
Mathematics
1 answer:
Art [367]3 years ago
4 0

Answer:

See below in bold.

Step-by-step explanation:

For the fair coin Prob(head) = 1/2 and Prob(Tail) = 1/2.

For the biased coin it is   Prob(head) = 2/3 and Prob(Tail) = 1/3.

a) Prob(2 heads) = 1/2 * 2/3 = 1/3.

b) Prob(2 tails) = 1/2 * 1/3 = 1/6.

c) Prob(1 head ) = Prob(H T or T H) = 1/2 * 1/3  + 1/2 * 2/3) = 1/6+1/3 = 1/2.

d) Prob (at least one head) = prob (HH or TH or HT) =  1/3 + 1/2 =<em> </em>5/6.

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Answer:

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Step-by-step explanation:

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The population of a colony of mosquitoes obeys the law of uninhibited growth. If there are 1000 mosquitoes initially and there a
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Answer:

1) The size of the colony after 4 ​days is 6553 mosquitoes

2) After t = 4.9\ days

Step-by-step explanation:

To answer this question you must use the growth formula

N = N_0e ^ {kt}

Where

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t is the time in days

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We know that when t = 1 and N_0=1000 then N = 1600

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Now that we know k we can find the size of the colony after 4 days.

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10000 = 1000e ^ {0.47t}\\\\10 = e ^ {0.47t}\\\\ln(10) = 0.47t\\\\t = \frac{ln(10)}{0.47}\\\\t = 4.9\ days

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