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Zorluk: KolayProbability of Independent, Dependent, and Mutually Exclusive Events

A fair six-sided die with faces numbered 11 through 66 is rolled twice. What is the probability of rolling a 44 on the first roll and an odd number on the second roll?

  1. 112\frac{1}{12}Cevap
  2. B
    23\frac{2}{3}
  3. C
    13\frac{1}{3}
  4. D
    14\frac{1}{4}
  5. E
    136\frac{1}{36}

Cevap

112\frac{1}{12}
Because the outcome of the first die roll does not affect the outcome of the second die roll, the two events are independent. The probability of rolling a 44 on the first roll is 16\frac{1}{6}, and the probability of rolling an odd number (1,3,1, 3, or 55) on the second roll is 36=12\frac{3}{6} = \frac{1}{2}. Multiplying these individual probabilities together yields 16×12=112\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}.

Adım Adım Çözüm

1
Determine the probability of the first event (rolling a 4).
There is 11 favorable outcome out of 66 possible outcomes, so P(First roll is 4)=16P(\text{First roll is } 4) = \frac{1}{6}.
Each face of a fair six-sided die is equally likely to land face up.
2
Determine the probability of the second event (rolling an odd number).
The odd numbers are 1,3,1, 3, and 55, giving 33 favorable outcomes out of 66 total outcomes, so P(Second roll is odd)=36=12P(\text{Second roll is odd}) = \frac{3}{6} = \frac{1}{2}.
Half of the outcomes on a standard six-sided die are odd.
3
Calculate the joint probability of both independent events occurring.
P(4 on first AND odd on second)=16×12=112P(\text{4 on first AND odd on second}) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}.
For independent events AA and BB, the multiplication rule states that P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B).

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Probability of Independent Events
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