Question

A random experiment was conducted where a Person A tossed five coins and recorded the number of ""heads"". Person B rolled two dice and recorded the larger number out of the two dice. Simulate this scenario (use 10000 long columns) and answer questions 10 to 13.

Answer 1

(10) Person B

(11) Person B

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

(13) Person B

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) = (n(5))/(n(Head))`

`Pr(5) = (1)/(32)`

`Pr(5) = 0.03125`

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

`Pr(5) = (n(5))/(n(Dice))`

`Pr(5) = (8)/(30)`

`Pr(5) = 0.267`

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

Question 11: Person with Higher median

For person A

`Median = (n(Head) + 1)/(2)th`

`Median = (32 + 1)/(2)th`

`Median = (33)/(2)th`

`Median = 16.5th`

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

So,

`Median = (3+2)/(2)`

`Median = (5)/(2)`

`Median = 2.5`

For person B

`Median = (n(Dice) + 1)/(2)th`

`Median = (30 + 1)/(2)th`

`Median = (31)/(2)th`

`Median = 15.5th`

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

`Median = (5+5)/(2)`

`Median = (10)/(2)`

`Median = 5`

Person B has a greater median of 5

Question 12: Probability that B gets 5 or 6

This is calculated as:

`P(5\ or\ 6) = (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) = (n(5\ or\ 6))/(n(Dice))`

`P(5\ or\ 6) = (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) = (n(3\ or\ more))/(n(Head))`

`P(3\ or\ more) = (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) = (n(3\ or\ more))/(n(Dice))`

`P(3\ or\ more) = (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