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nuccioreggie
 one year ago
help
nuccioreggie
 one year ago
help

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nuccioreggie
 one year ago
Best ResponseYou've already chosen the best response.0Jack uses a probability simulator to roll a sixsided 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here we have two independent events, namely the flipping of a coin and the rolling of a sixsided 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)=...?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please you have to compute p1 and p2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large \begin{gathered} {p_1} = \frac{{27}}{{16 + 14 + 5 + 17 + 21 + 27}} = ...? \hfill \\ \hfill \\ {p_2} = \frac{{41}}{{41 + 59}} = ...? \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I'm sorry, option D is incorrect!

nuccioreggie
 one year ago
Best ResponseYou've already chosen the best response.0oh wait sorry its c

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please 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?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, since those two ebents are independent events, then the requested probability is given by the subsequent product: \[\Large {p_1} \cdot {p_2} = \frac{{27}}{{100}} \cdot \frac{{41}}{{100}} = ...?\]
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