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Find the probability that Mensah passes both Mathematics and Economics, given the individual probabilities and the probability of passing at least one.

Mathematics

ScanSolve AI Answer

Answer

1924\frac{19}{24}

Step-by-step solution

5 steps
  1. Identify the given probabilities.

    Let P(M)P(M) be the probability that Mensah passes Mathematics.
    Let P(E)P(E) be the probability that Mensah passes Economics.
    Let P(M∪E)P(M \cup E) be the probability that Mensah passes at least one subject.
    We are given:
    P(M)=34P(M) = \frac{3}{4}
    P(E)=58P(E) = \frac{5}{8}
    P(M∪E)=712P(M \cup E) = \frac{7}{12}

  2. State the formula for the probability of the union of two events.

    The formula for the probability of the union of two events is:

    P(M∪E)=P(M)+P(E)−P(M∩E)P(M \cup E) = P(M) + P(E) - P(M \cap E)

    where P(M∩E)P(M \cap E) is the probability that Mensah passes both subjects.

  3. Rearrange the formula to solve for P(M∩E)P(M \cap E).

    P(M∩E)=P(M)+P(E)−P(M∪E)P(M \cap E) = P(M) + P(E) - P(M \cup E)
  4. Substitute the given values into the formula.

    P(M∩E)=34+58−712P(M \cap E) = \frac{3}{4} + \frac{5}{8} - \frac{7}{12}
  5. Find a common denominator for the fractions and perform the calculation.

    The least common multiple (LCM) of 4, 8, and 12 is 24.
    Convert each fraction to have a denominator of 24:
    34=3×64×6=1824\frac{3}{4} = \frac{3 \times 6}{4 \times 6} = \frac{18}{24}
    58=5×38×3=1524\frac{5}{8} = \frac{5 \times 3}{8 \times 3} = \frac{15}{24}
    712=7×212×2=1424\frac{7}{12} = \frac{7 \times 2}{12 \times 2} = \frac{14}{24}

    Now substitute these equivalent fractions back into the equation:

    P(M∩E)=1824+1524−1424P(M \cap E) = \frac{18}{24} + \frac{15}{24} - \frac{14}{24} P(M∩E)=18+15−1424P(M \cap E) = \frac{18 + 15 - 14}{24} P(M∩E)=33−1424P(M \cap E) = \frac{33 - 14}{24} P(M∩E)=1924P(M \cap E) = \frac{19}{24}

    The probability that Mensah passes both subjects is 1924\frac{19}{24}.

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