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BartSMP [9]
3 years ago
6

Sally has two coins. The first coin is a fair coin and the second coin is biased. The biased coin comes up heads with probabilit

y .75 and tails with probability .25. She selects a coin at random and flips the coin ten times. The results of the coin flips are mutually independent. The result of the 10 flips is: T,T,H,T,H,T,T,T,H,T. What is the probability that she selected the biased coin?
Mathematics
1 answer:
coldgirl [10]3 years ago
4 0

Answer:

50%

Step-by-step explanation:

''What is the probability that she selected the biased coin?”

When we have n possible outcomes of an event and all of them have the same probability of appearance, then in theory each possibility has a probability 1/n of being the result of the event.

In this case the event is choosing randomly a coin out of 2, so no matter what the biased coin is or the results we get when we toss it, the probability of choosing the biased coin is ½ = 0.5 or 50%.

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<em><u>Question:</u></em>

At the city museum, child admission is $5.80 and adult admission is $9.30. On Monday, four times as many adult tickets as child tickets were sold, for a total sales of $1548.00. How many child tickets were sold that day?

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<em><u>Solution:</u></em>

Given that,

Cost of 1 child admission = $ 5.80

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Let "c" be the number of child tickets sold

Let "a" be the number of adult tickets sold

On Monday, four times as many adult tickets as child tickets were sold

Number of adult tickets sold = four times the number of child tickets

Number of adult tickets sold = 4(number of child tickets sold)

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They were sold for a total sales of $ 1548.00

number of child tickets sold x Cost of 1 child admission + number of adult tickets sold x Cost of 1 adult admission = 1548.00

c \times 5.80 + a \times 9.30 = 1548

5.8c + 9.3a = 1548  ---- eqn 2

Let us solve eqn 1 and eqn 2 to find values of "c" and "a"

Substitute eqn 1 in eqn 2

5.8c + 9.3(4c) = 1548

5.8c + 37.2c = 1548

43c = 1548

c = 36

Thus 36 child tickets were sold that day

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