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If two fair coins are tossed, what is the probability of getting at least one head?

Mathematics

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Answer

B

Step-by-step solution

7 steps

Here are the solutions to the problems:

Part 11· 4 steps

  1. Determine the total number of possible outcomes.

    Bola chose a number between 1 and 300, inclusive.
    Total number of outcomes = 300.

  2. Determine the number of favorable outcomes (numbers divisible by 4).

    To find how many numbers between 1 and 300 are divisible by 4, divide 300 by 4.
    Number of favorable outcomes = 3004=75\frac{300}{4} = 75.

  3. Calculate the probability.

    P(divisible by 4)=Number of favorable outcomesTotal number of outcomesP(\text{divisible by 4}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} P(divisible by 4)=75300P(\text{divisible by 4}) = \frac{75}{300}
  4. Simplify the probability.

    P(divisible by 4)=14P(\text{divisible by 4}) = \frac{1}{4}

    The correct option is B.

    B. 14\boxed{\text{B. } \frac{1}{4}}

Part 12· 3 steps

  1. Identify the probabilities of each event.

    Let P(K)P(K) be the probability Mr. Kunle passes = 13\frac{1}{3}.
    Let P(B)P(B) be the probability Mr. Boye passes = 23\frac{2}{3}.

  2. Calculate the probability that neither passes.

    The probability Mr. Kunle fails is P(K′)=1−P(K)=1−13=23P(K') = 1 - P(K) = 1 - \frac{1}{3} = \frac{2}{3}.
    The probability Mr. Boye fails is P(B′)=1−P(B)=1−23=13P(B') = 1 - P(B) = 1 - \frac{2}{3} = \frac{1}{3}.
    Assuming the events are independent, the probability that neither passes is P(K′ and B′)=P(K′)×P(B′)P(K' \text{ and } B') = P(K') \times P(B').

    P(neither passes)=23×13=29P(\text{neither passes}) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}
  3. Calculate the probability that at least one of them passes.

    The probability that at least one passes is the complement of neither passing.

    P(at least one passes)=1−P(neither passes)P(\text{at least one passes}) = 1 - P(\text{neither passes}) P(at least one passes)=1−29P(\text{at least one passes}) = 1 - \frac{2}{9} P(at least one passes)=99−29P(\text{at least one passes}) = \frac{9}{9} - \frac{2}{9} P(at least one passes)=79P(\text{at least one passes}) = \frac{7}{9}

    The correct option is A (or C).

    A. 79\boxed{\text{A. } \frac{7}{9}}
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