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13 A fair coin is tossed 3 times, what is the probability of getting at least two heads? A. 2/3 B. 4/5 C. 2/5 D. 1/2

Mathematics

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Part 12· 3 steps

  1. Identify the probabilities of each person passing.

    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}.
    Since the events are independent, the probability that neither passes is the product of their individual failure probabilities.

    P(neither passes)=P(K′)×P(B′)=23×13=29P(\text{neither passes}) = P(K') \times P(B') = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}

  3. Calculate the probability that at least one of them passes.

    The event "at least one passes" is the complement of the event "neither passes".

    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.

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

Part 13· 4 steps

  1. List all possible outcomes when a fair coin is tossed 3 times.

    The total number of outcomes is 23=82^3 = 8.
    The outcomes are: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.

  2. Identify the outcomes with at least two heads.

    "At least two heads" means two heads or three heads.
    The favorable outcomes are: HHH, HHT, HTH, THH.
    There are 4 favorable outcomes.

  3. Calculate the probability.

    P(at least two heads)=Number of favorable outcomesTotal number of outcomesP(\text{at least two heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} P(at least two heads)=48P(\text{at least two heads}) = \frac{4}{8}

  4. Simplify the probability.

    P(at least two heads)=12P(\text{at least two heads}) = \frac{1}{2}

    The correct option is D.

    D. 12\boxed{\text{D. } \frac{1}{2}}

Part 14· 4 steps

  1. Determine the total number of outcomes when two fair dice are rolled.

    Each die has 6 faces, so for two dice, the total number of outcomes is 6×6=366 \times 6 = 36.

  2. Identify the outcomes where both dice show the same number.

    The favorable outcomes are: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).
    There are 6 favorable outcomes.

  3. Calculate the probability.

    P(same number)=Number of favorable outcomesTotal number of outcomesP(\text{same number}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} P(same number)=636P(\text{same number}) = \frac{6}{36}

  4. Simplify the probability.

    P(same number)=16P(\text{same number}) = \frac{1}{6}

    The correct option is D.

    D. 16\boxed{\text{D. } \frac{1}{6}}

Part 15· 3 steps

  1. List all possible outcomes when two fair coins are tossed.

    The total number of outcomes is 22=42^2 = 4.
    The outcomes are: HH, HT, TH, TT.

  2. Identify the outcomes with at least one head.

    "At least one head" means one head or two heads.
    The favorable outcomes are: HH, HT, TH.
    There are 3 favorable outcomes.

  3. Calculate the probability.

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

    The correct option is D.

    D. 34\boxed{\text{D. } \frac{3}{4}}

Part 16· 4 steps

  1. Determine the total number of outcomes when two dice are thrown together.

    The total number of outcomes is 6×6=366 \times 6 = 36.

  2. Identify the outcomes where the sum of the numbers is 10.

    The favorable outcomes are: (4,6), (5,5), (6,4).
    There are 3 favorable outcomes.

  3. Calculate the probability.

    P(sum of 10)=Number of favorable outcomesTotal number of outcomesP(\text{sum of 10}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} P(sum of 10)=336P(\text{sum of 10}) = \frac{3}{36}

  4. Simplify the probability.

    P(sum of 10)=112P(\text{sum of 10}) = \frac{1}{12}

    The correct option is B.

    B. 112\boxed{\text{B. } \frac{1}{12}}

Part 17· 4 steps

  1. Determine the probability of the top face being white on a single throw.

    A die has 6 faces. 4 faces are white.

    P(White on one throw)=Number of white facesTotal number of faces=46P(\text{White on one throw}) = \frac{\text{Number of white faces}}{\text{Total number of faces}} = \frac{4}{6}

  2. Simplify the probability for a single throw.

    P(White on one throw)=23P(\text{White on one throw}) = \frac{2}{3}

  3. Calculate the probability of the top face being white in two consecutive throws.

    Since the throws are independent events, multiply their probabilities.

    P(White in both cases)=P(White on 1st throw)×P(White on 2nd throw)P(\text{White in both cases}) = P(\text{White on 1st throw}) \times P(\text{White on 2nd throw}) P(White in both cases)=23×23P(\text{White in both cases}) = \frac{2}{3} \times \frac{2}{3}

  4. Calculate the final probability.

    P(White in both cases)=49P(\text{White in both cases}) = \frac{4}{9}

    The correct option is C.

    C. 49\boxed{\text{C. } \frac{4}{9}}
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