## nuccioreggie one year ago help

1. nuccioreggie

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

2. nuccioreggie

@Michele_Laino

3. Michele_Laino

here we have two independent events, namely the flipping of a coin and the rolling of a six-sided cube. The probability to get a six, is: p1=favorable outcomes / possible outcomes= =27/(16+14+5+17+21+27)=...? the probability to get a head is: p2=favorable outcomes/ possible outcomes= 41/(41+59)=...?

4. nuccioreggie

60

5. Michele_Laino

what is 60?

6. Michele_Laino

please you have to compute p1 and p2

7. nuccioreggie

95.6875

8. Michele_Laino

$\Large \begin{gathered} {p_1} = \frac{{27}}{{16 + 14 + 5 + 17 + 21 + 27}} = ...? \hfill \\ \hfill \\ {p_2} = \frac{{41}}{{41 + 59}} = ...? \hfill \\ \end{gathered}$

9. nuccioreggie

so D

10. Michele_Laino

I'm sorry, option D is incorrect!

11. nuccioreggie

oh wait sorry its c

12. Michele_Laino

please check my computation: $\Large \begin{gathered} {p_1} = \frac{{27}}{{16 + 14 + 5 + 17 + 21 + 27}} = \frac{{27}}{{100}} \hfill \\ \hfill \\ {p_2} = \frac{{41}}{{41 + 59}} = \frac{{41}}{{100}} \hfill \\ \end{gathered}$ am I right?

13. nuccioreggie

yes so its C

14. Michele_Laino

now, since those two ebents are independent events, then the requested probability is given by the subsequent product: $\Large {p_1} \cdot {p_2} = \frac{{27}}{{100}} \cdot \frac{{41}}{{100}} = ...?$

15. Michele_Laino

events*

16. nuccioreggie

A

17. Michele_Laino

that's right!