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madam [21]
3 years ago
7

When a fair coin is flipped, what is the chance of getting about 50% heads- specifically between 40% and 60% heads for n=10 flip

s?
Mathematics
1 answer:
aleksklad [387]3 years ago
3 0
At at least one die come up a 3?We can do this two ways:) The straightforward way is as follows. To get at least one 3, would be consistent with the following three mutually exclusive outcomes:the 1st die is a 3 and the 2nd is not: prob = (1/6)x(5/6)=5/36the 1st die is not a 3 and the 2nd is: prob = (5/6)x((1/6)=5/36both the 1st and 2nd come up 3: prob = (1/6)x(1/6)=1/36sum of the above three cases is prob for at least one 3, p = 11/36ii) A faster way is as follows: prob at least one 3 = 1 - (prob no 3's)The probability to get no 3's is (5/6)x(5/6) = 25/36.So the probability to get at least one 3 is, p = 1 - (25/36) = 11/362) What is the probability that a card drawn at random from an ordinary 52 deck of playing cards is a queen or a heart?There are 4 queens and 13 hearts, so the probability to draw a queen is4/52 and the probability to draw a heart is 13/52. But the probability to draw a queen or a heart is NOT the sum 4/52 + 13/52. This is because drawing a queen and drawing a heart are not mutually exclusive outcomes - the queen of hearts can meet both criteria! The number of cards which meet the criteria of being either a queen or a heart is only 16 - the 4 queens and the 12 remaining hearts which are not a queen. So the probability to draw a queen or a heart is 16/52 = 4/13.3) Five coins are tossed. What is the probability that the number of heads exceeds the number of tails?We can divide
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<h3>Answer:  9</h3>

================================================

Work Shown:

a = unknown  = leg #1

b = 12  = leg #2

c = 15 = hypotenuse

Plug those values into the pythagorean theorem and solve for 'a'

a^2 + b^2 = c^2

a^2 + (12)^2 = (15)^2  .... substitution

a^2 + 144 = 225

a^2 = 225 - 144  ... subtracting 144 from both sides

a^2 = 81

a = sqrt(81)  .... applying square root to both sides

a = 9

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