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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
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)=...?

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Other answers:

60
what is 60?
please you have to compute p1 and p2
95.6875
\[\Large \begin{gathered} {p_1} = \frac{{27}}{{16 + 14 + 5 + 17 + 21 + 27}} = ...? \hfill \\ \hfill \\ {p_2} = \frac{{41}}{{41 + 59}} = ...? \hfill \\ \end{gathered} \]
so D
I'm sorry, option D is incorrect!
oh wait sorry its c
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?
yes so its C
now, since those two ebents are independent events, then the requested probability is given by the subsequent product: \[\Large {p_1} \cdot {p_2} = \frac{{27}}{{100}} \cdot \frac{{41}}{{100}} = ...?\]
events*
A
that's right!

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