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Assumption: The probability that Forson will pass the examination is (3)/(4), as the image cuts off the last fraction.

Mathematics

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3 parts
i

516\frac{5}{16}

ii

132\frac{1}{32}

iii

512\frac{5}{12}

24

Step-by-step solution

13 steps

Here is the solution to your question.

Assumption: The probability that Forson will pass the examination is 34\frac{3}{4}, as the image cuts off the last fraction.

Part 9(a): Probability

Let P(A)P(A), P(I)P(I), and P(F)P(F) be the probabilities that Ayuba, Idris, and Forson pass the examination, respectively.
Given:
P(A)=23P(A) = \frac{2}{3}
P(I)=58P(I) = \frac{5}{8}
P(F)=34P(F) = \frac{3}{4}

The probabilities that they fail are:
P(A′)=1−P(A)=1−23=13P(A') = 1 - P(A) = 1 - \frac{2}{3} = \frac{1}{3}
P(I′)=1−P(I)=1−58=38P(I') = 1 - P(I) = 1 - \frac{5}{8} = \frac{3}{8}
P(F′)=1−P(F)=1−34=14P(F') = 1 - P(F) = 1 - \frac{3}{4} = \frac{1}{4}

Part i· 3 steps

The three of them will pass the examination

  1. Multiply the individual probabilities of passing.

    P(A and I and F pass)=P(A)×P(I)×P(F)P(\text{A and I and F pass}) = P(A) \times P(I) \times P(F) P(A and I and F pass)=23×58×34P(\text{A and I and F pass}) = \frac{2}{3} \times \frac{5}{8} \times \frac{3}{4}
  2. Calculate the product.

    P(A and I and F pass)=2×5×33×8×4=3096P(\text{A and I and F pass}) = \frac{2 \times 5 \times 3}{3 \times 8 \times 4} = \frac{30}{96}
  3. Simplify the fraction.

    P(A and I and F pass)=516P(\text{A and I and F pass}) = \frac{5}{16}

    The probability that the three of them will pass is 516\boxed{\frac{5}{16}}.

Part ii

132\frac{1}{32}

None of them will pass the examination

  1. Multiply the individual probabilities of failing.

    P(None pass)=P(A′)×P(I′)×P(F′)P(\text{None pass}) = P(A') \times P(I') \times P(F') P(None pass)=13×38×14P(\text{None pass}) = \frac{1}{3} \times \frac{3}{8} \times \frac{1}{4}
  2. Calculate the product.

    P(None pass)=1×3×13×8×4=396P(\text{None pass}) = \frac{1 \times 3 \times 1}{3 \times 8 \times 4} = \frac{3}{96}
  3. Simplify the fraction.

    P(None pass)=132P(\text{None pass}) = \frac{1}{32}

    The probability that none of them will pass is 132\boxed{\frac{1}{32}}.

Part iii

512\frac{5}{12}

Ayuba and Idris will pass the examination

  1. Multiply the probabilities of Ayuba and Idris passing.

    P(A and I pass)=P(A)×P(I)P(\text{A and I pass}) = P(A) \times P(I) P(A and I pass)=23×58P(\text{A and I pass}) = \frac{2}{3} \times \frac{5}{8}
  2. Calculate the product.

    P(A and I pass)=1024P(\text{A and I pass}) = \frac{10}{24}
  3. Simplify the fraction.

    P(A and I pass)=512P(\text{A and I pass}) = \frac{5}{12}

    The probability that Ayuba and Idris will pass is 512\boxed{\frac{5}{12}}.

    Part 9(b): Ratio Problem

  4. Find the initial number of boys and girls.

    Total students = 432
    Ratio of boys to girls = 5:4
    Total ratio parts = 5+4=95 + 4 = 9

    Initial number of boys:

    Binitial=59×432=5×48=240B_{initial} = \frac{5}{9} \times 432 = 5 \times 48 = 240

    Initial number of girls:

    Ginitial=49×432=4×48=192G_{initial} = \frac{4}{9} \times 432 = 4 \times 48 = 192
  5. Calculate the new number of boys.

    The number of boys increases by 12.

    Bnew=Binitial+12=240+12=252B_{new} = B_{initial} + 12 = 240 + 12 = 252
  6. Use the new ratio to find the new number of girls.

    The new ratio of boys to girls is 7:6.

    BnewGnew=76\frac{B_{new}}{G_{new}} = \frac{7}{6} 252Gnew=76\frac{252}{G_{new}} = \frac{7}{6} Gnew=252×67G_{new} = \frac{252 \times 6}{7} Gnew=36×6=216G_{new} = 36 \times 6 = 216
  7. Find the increase in the number of girls.

    Increase in girls = Gnew−GinitialG_{new} - G_{initial}
    Increase in girls = 216−192=24216 - 192 = 24

    The increase in the number of girls is 24\boxed{24}.

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