A die has one red face, two blue faces, and three green faces. It is rolled 5 times. Find the chance that the red face appears on one of the rolls and the remaining rolls are green.
[Careful what you multiply. The most straightforward method is to follow the derivation of the binomial formula.]
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
(1/6)(3/6)^4 = ?
0.052
C(5,1)*(1/6)*(3/6)^4=0.05208333
Thanks friends, correct
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You guys crack me up.
0.0287
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Nolan concha do you have?
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To find the chance of the red face appearing on exactly one roll and the remaining rolls being green, we need to consider the total number of possible outcomes and the number of favorable outcomes.
Total number of possible outcomes = Total number of ways to roll the die 5 times
= 6 options (red, blue1, blue2, green1, green2, green3) for each roll, and since there are 5 rolls, the total number of possible outcomes is 6^5.
Number of favorable outcomes = Number of ways the red face appears on one roll and the remaining rolls are green
= 1 way for the red face to appear on one roll (red, green, green, green, green), and for the remaining 4 rolls, there are 3 options (green1, green2, green3) each time. So the total number of favorable outcomes is 1 * 3^4.
Now, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = Number of Favorable Outcomes / Total Number of Possible Outcomes
= (1 * 3^4) / 6^5
Simplifying this expression, we get:
Probability = (1 * 81) / 7776
= 81 / 7776
= 1 / 96
Therefore, the chance that the red face appears on one of the rolls and the remaining rolls are green is 1/96.