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Helen [10]
3 years ago
12

In a game, a participant is given three attempts to hit a ball. On each try, she either scores a hit, H, or a miss. M. The game

requires that the player must alternate which hand she uses in successive attempts. That is, if she makes her first attempt with her right hand, she must use her left hand for the second attempt and her right hand for the third. Her chance of scoring a hit with her right hand is .7 and with her left hand is .4. Assume that the results of successive attempts are independent and that she wins the game if she scores at least two hits in a row. If she makes her first attempt with her right hand, what is the probability that she wins the game?
Mathematics
1 answer:
professor190 [17]3 years ago
6 0

Answer:

The probability that she wins the game is 0.364

Step-by-step explanation:

Let H = Hit

Let M = Miss

P(Hit with right hand) = 0.7

P(Hit with right hand) = 1-0.3 = 0.3

P(Hit with left hand) = 0.4

P(Miss with left hand) = 1-0.4 = 0.6

First, we need to highlight possible outcomes.

Let SS = Sample Space

SS = {HHH, HHM, HMH, MHH, HMM, MHM, MMH, MMM}

At this point, we follow the assumption that she starts with her right hand (according to the question)

Out of the possible events, only 3 will have the participant win the game:

Which are:

HHH, HHM and MHH.

P (HHH) + P (HHM) + P (MHH)

P(HHH) = P(Hit with right) and P(Hit with left) and P(Hit with right)

P(HHH) = 0.7 * 0.4 * 0.7

P(HHH) = 0.196

P(HHM) = P(Hit with right) + P(Hit with left) + P(Miss with right)

P(HHM) = 0.7 * 0.4 * 0.3

P(HHM) = 0.084

P(MHH) = 0.084 (Same as above)

Probability that she wins = P (HHH) + P (HHM) + P (MHH) = 0.196 + 0.084 + 0.084

Probability that she wins = 0.364

..

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Answer:

(a) Margin of error ( E) = $2,000 , n = 54

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Step-by-step explanation:

Given -

Standard deviation \sigma = $7,500

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Z_{\frac{\alpha}{2}} =  Z_{\frac{.05}{2}} = 1.96

let sample size is n

(a) Margin of error ( E) = $2,000

Margin of error ( E)  = Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}

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Squaring both side

E^{2} = 1.96^{2}\times\frac{7500^{2}}{n}

n =\frac{1.96^{2}}{2000^{2}} \times 7500^{2}

n =  54.0225

n = 54 ( approximately)

(b)   Margin of error ( E) = $1,000

          E     = Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}

         1000   =  Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}

Squaring both side

1000^{2} = 1.96^{2}\times\frac{7500^{2}}{n}

n =\frac{1.96^{2}}{1000^{2}} \times 7500^{2}

n = 216

(c)   Margin of error ( E) = $500

   E = Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}

  500 = Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}

Squaring both side

500^{2} = 1.96^{2}\times\frac{7500^{2}}{n}

n =\frac{1.96^{2}}{500^{2}} \times 7500^{2}

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Step-by-step explanation:

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For the given material that has a mass of 44 grams occupying a volume of 6 cm^3, we divide the mass by the volume, i.e.

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Step-by-step explanation:

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