Question

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

Answer 1

Answer:

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.  $\dfrac{1107}{10000}$      

Step-by-step explanation:

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,

$P(A\bigcap B)=P(A)\times P(B)$

Now, based on the two tables we have:

$P(A)=\dfrac{27}{100}$

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

Also,

$P(B)=\dfrac{41}{100}$

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

Hence, we get:

$P(A\bigcap B)=\dfrac{27}{100}\times \dfrac{41}{100}$

i.e.

$P(A\bigcap B)=\dfrac{27\times 41}{100\times 100}$

i.e.

$P(A\bigcap B)=\dfrac{1107}{10000}$

Answer 2

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%)