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oksano4ka [1.4K]
3 years ago
6

A game at the fair involves a wheel with seven sectors. Two of the sectors are red, two of the sectors are purple, two of the se

ctors are yellow, and one sector is blue. Landing on the blue sector will give 3 points, landing on a yellow sector will give 1 point, landing on a purple sector will give 0 points, and landing on a red sector will give –1 point.
If you take one spin, what is your expected value?

What changes could you make to values assigned to outcomes to make the game fair? Prove that the game would be fair using expected values.
Mathematics
1 answer:
Zolol [24]3 years ago
7 0

Answer:

Expected value: 3/7

Change landing on purple to -1.5

Step-by-step explanation:

Expected value

=3(1/7) + 1(2/7) + 0(2/7) - 1(2/7) = 3/7

Fair game: Expected value = 0

Change the value of 0 to -1.5

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6x+5y+3x+7-5y Fully Simplify expression below​
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Answer:

9x + 7

Step-by-step explanation:

Given

6x + 5y + 3x + 7 - 5y ← collect like terms

= (6x + 3x ) + (5y - 5y ) + 7

= 9x + 0 + 7

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A pharmaceutical company receives large shipments of ibuprofen tablets and uses an acceptance sampling plan. This plan randomly
Keith_Richards [23]

Answer:

0.7208 = 72.08% probability that this whole shipment will be accepted.

Step-by-step explanation:

For each tablet, there are only two possible outcomes. Either it meets the required specifications, or it does not. The probability of a tablet meeting the required specifications is independent of any other tablet, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

4% rate of defects

This means that p = 0.04

26 tablets

This means that n = 26

What is the probability that this whole shipment will be accepted?

Probability that at most one tablet does not meet the specifications, which is:

P(X \leq 1) = P(X = 0) + P(X = 1)

Thus

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{26,0}.(0.04)^{0}.(0.96)^{26} = 0.3460

P(X = 1) = C_{26,1}.(0.04)^{1}.(0.96)^{25} = 0.3748

Then

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.3460 + 0.3748 = 0.7208

0.7208 = 72.08% probability that this whole shipment will be accepted.

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