Jack uses a probability simulator to roll a six-sided number cube 100 times and to flip a coin 100 times. The results of the experiment are shown below:

Number on the Cube Number of Times Rolled
1 16
2 14
3 5
4 17
5 21
6 27

Heads Tails
41 59

Using Jack's simulation, what is the probability of rolling a 6 on the number cube and the coin landing on heads?
fraction 1,107 over 10,000
fraction 1,593 over 10,000
fraction 27 over 100
fraction 41 over 100

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AceTalon21 AceTalon21 · Answer 1 · 2018-09-07T17:56:13+02:00

The experimental probability is the number of specific outcomes divided by the sample size...

P(6)=27/100 (27%)

P(H)=41/100 (41%)

Not sure, but if you meant rolling a 6 AND getting a head then:

P(6 AND H)=(27/100)(41/100)=1107/10000 (11.07%)

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lidaralbany lidaralbany · Answer 2 · 2019-07-20T03:53:34+02:00

The probability of rolling a 6 on the number cube and the coin landing on heads is:

fraction 1,107 over 10,000 i.e. [tex]\dfrac{1107}{10000}[/tex]

Let A denote the event of rolling a 6 on number cube.

and B denote the event of landing a head on a coin.

Clearly both the events A and B are independent.

Also, let P denote the probability of an event.

We are asked to find: P(A∩B)

We know that when two events A and B are independent.

Then,

[tex]P(A\bigcap B)=P(A)\times P(B)[/tex]

Now, based on the two tables we have:

[tex]P(A)=\dfrac{27}{100}[/tex]

( Since, 6 comes up on rolling a number cube 27 times out of a total of 100 times)

Also,

[tex]P(B)=\dfrac{41}{100}[/tex]

( since head comes up 41 times out of a total of 100 times)

Hence, we get:

[tex]P(A\bigcap B)=\dfrac{27}{100}\times \dfrac{41}{100}[/tex]

i.e.

[tex]P(A\bigcap B)=\dfrac{27\times 41}{100\times 100}[/tex]

i.e.

[tex]P(A\bigcap B)=\dfrac{1107}{10000}[/tex]