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nadezda [96]
3 years ago
6

Maria believes she will get 1 hit in 20% of her softball games, 2 hits in 25% of her games, 3 hits in 50% of her games, and 4 hi

ts in 5% of her games. About how many hits can Maria expect to have each game?
1
2
3
4
Mathematics
2 answers:
castortr0y [4]3 years ago
5 0

Answer:

<em>The number of hits can Maria expect to have each game is </em><em>2 </em>

Step-by-step explanation:

The expected value is the sum of product of each of the possible outcomes and the probability of the outcome occurring. Mathematically,

E(X) =\sum x\cdot P(X = x)

Given,

P(X=1)=0.20\\\\P(X=2)=0.25\\\\P(X=3)=0.50\\\\P(X=4)=0.05

Putting the values,

E(X)=(1\times 0.20)+(2\times 0.25)+(3\times 0.50)+(4\times 0.05)=2.4

As the number of hits can not be in fraction, so the number of hits can Maria expect to have each game is 2

creativ13 [48]3 years ago
3 0
I think 3 hits as the percentage is higher.
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7 0
3 years ago
Drone INC. owns four 3D printers (D1, D2, D3, D4) that print all their Drone parts. Sometimes errors in printing occur. We know
USPshnik [31]

Answer:

Step-by-step explanation:

Hello!

There are 4 3D printers available to print drone parts, then be "Di" the event that the printer i printed the drone part (∀ i= 1,2,3,4), and the probability of a randomly selected par being print by one of them is:

D1 ⇒ P(D1)= 0.15

D2 ⇒ P(D2)= 0.25

D3 ⇒ P(D3)= 0.40

D4 ⇒ P(D4)= 0.20

Additionally, there is a chance that these printers will print defective parts. Be "Ei" represent the error rate of each print (∀ i= 1,2,3,4) then:

P(E1)= 0.01

P(E2)= 0.02

P(E3)= 0.01

P(E4)= 0.02

Ei is then the event that "the piece was printed by Di" and "the piece is defective".

You need to determine the probability of randomly selecting a defective part printed by each one of these printers, i.e. you need to find the probability of the part being printed by the i printer given that is defective, symbolically: P(DiIE)

Where "E" represents the event "the piece is defective" and its probability represents the total error rate of the production line:

P(E)= P(E1)+P(E2)+P(E3)+P(E4)= 0.01+0.02+0.01+0.02= 0.06

This is a conditional probability and you can calculate it as:

P(A/B)= \frac{P(AnB)}{P(B)}

To reach the asked probability, first, you need to calculate the probability of the intersection between the two events, that is, the probability of the piece being printed by the Di printer and being defective Ei.

P(D1∩E)= P(E1)= 0.01

P(D2∩E)= P(E2)= 0.02

P(D3∩E)= P(E3)= 0.01

P(D4∩E)= P(E4)= 0.02

Now you can calculate the probability of the piece bein printed by each printer given that it is defective:

P(D1/E)= \frac{P(E1)}{P(E)} = \frac{0.01}{0.06}= 0.17

P(D2/E)= \frac{P(E2)}{P(E)} = \frac{0.02}{0.06}= 0.33

P(D3/E)= \frac{P(E3)}{P(E)} = \frac{0.01}{0.06}= 0.17

P(D4/E)= \frac{P(E4)}{P(E)} = \frac{0.02}{0.06}= 0.33

P(D2)= 0.25 and P(D2/E)= 0.33 ⇒ The prior probability of D2 is smaller than the posterior probability.

The fact that P(D2) ≠ P(D2/E) means that both events are nor independent and the occurrence of the piece bein defective modifies the probability of it being printed by the second printer (D2)

I hope this helps!

8 0
3 years ago
Rewrite the fraction 29/50 with a denominator of 100
romanna [79]

Answer:

58/100

50 X 2 = 100

29 X 2 = 58

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hodyreva [135]

Answer:

3/6 is equal

Step-by-step explanation:

because, if you had 3/6 of a pizza, you would have half of a pizza

8 0
2 years ago
Read 2 more answers
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