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SVEN [57.7K]
3 years ago
15

Find the mean:31, 1, 12, 2, 21, 22, 1, 0, 31, 13

Mathematics
1 answer:
Bond [772]3 years ago
5 0
The mean is 13.4 because after I add the numbers in the set, the sum comes out to be 134. There are 10 numbers in the set, so you divide 134 by 10, which is 13.4 (134 <span>÷ 10 = 13.4). That is how you get the mean. It's "mean" to you because you 'gotta do a lot of work to get your final mean. </span>
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A gambler has a fair coin and a two-headed coin in his pocket. He selects one of the coins at random. (a) When he flips the coin
Aleks04 [339]

Answer:

a) probability of choosing heads= 1/2 (50%)

b)  probability of choosing the fair coin knowing that it showed heads is= 1/3 (33.33%)

Step-by-step explanation:

Since the unfair coin can have 2 heads or 2 tails , and assuming both are equally possible . then

probability of choosing the fair coin  (named A)= 1/2

probability of choosing an unfair coin with 2 heads (named B)= (1-1/2)*1/2= 1/4

probability of choosing an unfair coin with 2 tails (named C)= (1-1/2)*(1-1/2)= 1/4

then

probability of choosing heads= probability of choosing A * probability of getting heads from A + probability of choosing B * probability of getting heads from B + probability of choosing C * probability of getting heads from C =

1/2*1/2 + 1/4*1 + 1/4*0 = 2/4 = 1/2

the probability of choosing the fair coin knowing that it showed heads is

P(A/B) = P(A∩B)/P(B)

denoting event A= the coin is fair and event B= the result is heads

P(A∩B) = 1/2*1/2 = 1/4

but since we know now that that the unfair coin is not possible , the probability of choosing heads is altered:

P(B)=probability of choosing heads= probability of choosing A * probability of getting heads from A + probability of choosing B * probability of getting heads from B  = 1/2*1/2+1/2*1 = 3/4

then

P(A/B) = P(A∩B)/P(B)  = (1/4)/(3/4) = 1/3

then the probability is 1/3

8 0
3 years ago
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