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9966 [12]
2 years ago
7

A random experiment was conducted where a Person A tossed five coins and recorded the number of ""heads"". Person B rolled two d

ice and recorded the larger number out of the two dice. Simulate this scenario (use 10000 long columns) and answer questions 10 to 13.
Mathematics
1 answer:
cestrela7 [59]2 years ago
7 0

Answer:

(10) Person B

(11) Person B

(12) P(5\ or\ 6) = 60\%

(13) Person B

Step-by-step explanation:

Given

Person A \to 5 coins (records the outcome of Heads)

Person \to Rolls 2 dice (recorded the larger number)

Person A

First, we list out the sample space of roll of 5 coins (It is too long, so I added it as an attachment)

Next, we list out all number of heads in each roll (sorted)

Head = \{5,4,4,4,4,4,3,3,3,3,3,3,3,3,3,3,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,0\}

n(Head) = 32

Person B

First, we list out the sample space of toss of 2 coins (It is too long, so I added it as an attachment)

Next, we list out the highest in each toss (sorted)

Dice = \{2,2,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6\}

n(Dice) = 30

Question 10: Who is likely to get number 5

From person A list of outcomes, the proportion of 5 is:

Pr(5) = \frac{n(5)}{n(Head)}

Pr(5) = \frac{1}{32}

Pr(5) = 0.03125

From person B list of outcomes, the proportion of 5 is:

Pr(5) = \frac{n(5)}{n(Dice)}

Pr(5) = \frac{8}{30}

Pr(5) = 0.267

<em>From the above calculations: </em>0.267 > 0.03125<em> Hence, person B is more likely to get 5</em>

Question 11: Person with Higher median

For person A

Median = \frac{n(Head) + 1}{2}th

Median = \frac{32 + 1}{2}th

Median = \frac{33}{2}th

Median = 16.5th

This means that the median is the mean of the 16th and the 17th item

So,

Median = \frac{3+2}{2}

Median = \frac{5}{2}

Median = 2.5

For person B

Median = \frac{n(Dice) + 1}{2}th

Median = \frac{30 + 1}{2}th

Median = \frac{31}{2}th

Median = 15.5th

This means that the median is the mean of the 15th and the 16th item. So,

Median = \frac{5+5}{2}

Median = \frac{10}{2}

Median = 5

<em>Person B has a greater median of 5</em>

Question 12: Probability that B gets 5 or 6

This is calculated as:

P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}

From the sample space of person B, we have:

n(5\ or\ 6) =n(5) + n(6)

n(5\ or\ 6) =8+10

n(5\ or\ 6) = 18

So, we have:

P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}

P(5\ or\ 6) = \frac{18}{30}

P(5\ or\ 6) = 0.60

P(5\ or\ 6) = 60\%

Question 13: Person with higher probability of 3 or more

Person A

n(3\ or\ more) = 16

So:

P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Head)}

P(3\ or\ more) = \frac{16}{32}

P(3\ or\ more) = 0.50

P(3\ or\ more) = 50\%

Person B

n(3\ or\ more) = 28

So:

P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Dice)}

P(3\ or\ more) = \frac{28}{30}

P(3\ or\ more) = 0.933

P(3\ or\ more) = 93.3\%

By comparison:

93.3\% > 50\%

Hence, person B has a higher probability of 3 or more

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<u>Step-by-step explanation:</u>

Here we have , A card is drawn at random from a deck of 52 playing cards. We need to find that Find the probability that the card drawn is:

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A deck of standard 52 cards contain four aces. There are four kings in a standard deck of playing cards. So ,we need to find probability that card drawn is either a ace or king i.e.

probability = \frac{Number-of-(ace+king)-cards}{total-number-of-cards}

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  • a king or a diamond

A deck of standard 52 cards contain four kings. There are 13 Diamonds in a standard deck of playing cards. So ,we need to find probability that card drawn is either a ace or king i.e.

probability = \frac{Number-of-(ace+king)-cards}{total-number-of-cards}

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⇒ probability = \frac{(17)}{52}

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Therefore, probability that the card drawn is:

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  2. a king or a diamond = 0.3269
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If you're using the app, try seeing this answer through your browser:  brainly.com/question/3242555

——————————

Solve the trigonometric equation:

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\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}


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\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.


\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}


\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}


You can discard the negative value for  t. So the solution for  (ii)  is

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where  k  is an integer.


I hope this helps. =)

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