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A fair coin is tossed twice. Find the probability that: a) 2 heads appear, b) 2 tails appear, c) at least one head appears.

Mathematics

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Answers

3 parts
a

14\frac{1}{4}

b

14\frac{1}{4}

c

34\frac{3}{4}

Step-by-step solution

6 steps

When a fair coin is tossed twice, the total possible outcomes are:
HH (Head, Head)
HT (Head, Tail)
TH (Tail, Head)
TT (Tail, Tail)
There are 4 total possible outcomes.

Part a· 2 steps

Find the probability that 2 heads appear.

  1. Identify the favorable outcome.

    The outcome where 2 heads appear is HH.

  2. Calculate the probability.

    There is 1 favorable outcome out of 4 total outcomes.

    P(2 heads)=Number of favorable outcomesTotal number of possible outcomes=14P(\text{2 heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{4}

    The probability that 2 heads appear is 14\boxed{\frac{1}{4}}.

Part b

14\frac{1}{4}

Find the probability that 2 tails appear.

  1. Identify the favorable outcome.

    The outcome where 2 tails appear is TT.

  2. Calculate the probability.

    There is 1 favorable outcome out of 4 total outcomes.

    P(2 tails)=Number of favorable outcomesTotal number of possible outcomes=14P(\text{2 tails}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{4}

    The probability that 2 tails appear is 14\boxed{\frac{1}{4}}.

Part c

34\frac{3}{4}

Find the probability that at least one head appears.

  1. Identify the favorable outcomes.

    "At least one head" means one head or two heads. The outcomes are HH, HT, TH.

  2. Calculate the probability.

    There are 3 favorable outcomes out of 4 total outcomes.

    P(at least one head)=Number of favorable outcomesTotal number of possible outcomes=34P(\text{at least one head}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{3}{4}

    The probability that at least one head appears is 34\boxed{\frac{3}{4}}.

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