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In a Science class 2/3 of the class are boys and the rest are girls. 80% of the boys and 90% of the girls are right handed. The probability that the right handed student will break a test tube in any session is 1/10 and that for the left handed student is 3/10 regardless of whether boy or girl.

Mathematics

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Answers

4 parts
b(i)

16\frac{1}{6}

b(ii)

120\frac{1}{20}

b(iii)

112\frac{1}{12}

b(iv)

1315\frac{13}{15}

Step-by-step solution

13 steps

Here's the solution to the problem:

Let B be the event that a student is a boy, G be the event that a student is a girl.
Let R be the event that a student is right-handed, L be the event that a student is left-handed.
Let D be the event that a test tube is broken, D' be the event that a test tube is not broken.

Given probabilities:
P(B)=23P(B) = \frac{2}{3}
P(G)=1−P(B)=1−23=13P(G) = 1 - P(B) = 1 - \frac{2}{3} = \frac{1}{3}

For boys:
P(R∣B)=80%=80100=45P(R|B) = 80\% = \frac{80}{100} = \frac{4}{5}
P(L∣B)=1−P(R∣B)=1−45=15P(L|B) = 1 - P(R|B) = 1 - \frac{4}{5} = \frac{1}{5}

For girls:
P(R∣G)=90%=90100=910P(R|G) = 90\% = \frac{90}{100} = \frac{9}{10}
P(L∣G)=1−P(R∣G)=1−910=110P(L|G) = 1 - P(R|G) = 1 - \frac{9}{10} = \frac{1}{10}

For breaking a test tube:
P(D∣R)=110P(D|R) = \frac{1}{10}
P(D′∣R)=1−P(D∣R)=1−110=910P(D'|R) = 1 - P(D|R) = 1 - \frac{1}{10} = \frac{9}{10}
P(D∣L)=310P(D|L) = \frac{3}{10}
P(D′∣L)=1−P(D∣L)=1−310=710P(D'|L) = 1 - P(D|L) = 1 - \frac{3}{10} = \frac{7}{10}

Part a

Draw a tree diagram to represent this information

The tree diagram branches out as follows:
\begin{itemize}
\item First level (Gender):
\begin{itemize}
\item Boy (B) with probability 23\frac{2}{3}
\item Girl (G) with probability 13\frac{1}{3}
\end{itemize}
\item Second level (Handedness, conditional on Gender):
\begin{itemize}
\item From Boy (B):
\begin{itemize}
\item Right-handed (R) with probability 45\frac{4}{5}
\item Left-handed (L) with probability 15\frac{1}{5}
\end{itemize}
\item From Girl (G):
\begin{itemize}
\item Right-handed (R) with probability 910\frac{9}{10}
\item Left-handed (L) with probability 110\frac{1}{10}
\end{itemize}
\end{itemize}
\item Third level (Test tube status, conditional on Handedness):
\begin{itemize}
\item From Right-handed (R) (for both boys and girls):
\begin{itemize}
\item Broken (D) with probability 110\frac{1}{10}
\item Not Broken (D') with probability 910\frac{9}{10}
\end{itemize}
\item From Left-handed (L) (for both boys and girls):
\begin{itemize}
\item Broken (D) with probability 310\frac{3}{10}
\item Not Broken (D') with probability 710\frac{7}{10}
\end{itemize}
\end{itemize}
\end{itemize}

Part b(i)

16\frac{1}{6}

A student chosen at random from the class is left-handed

Using the tree diagram drawn, find the probability

  1. Identify the paths leading to a student being left-handed.

    These are (Boy and Left-handed) or (Girl and Left-handed).

  2. Calculate the probability for each path.

    P(B∩L)=P(B)×P(L∣B)=23×15=215P(B \cap L) = P(B) \times P(L|B) = \frac{2}{3} \times \frac{1}{5} = \frac{2}{15} P(G∩L)=P(G)×P(L∣G)=13×110=130P(G \cap L) = P(G) \times P(L|G) = \frac{1}{3} \times \frac{1}{10} = \frac{1}{30}

  3. Sum the probabilities of these mutually exclusive paths.

    P(L)=P(B∩L)+P(G∩L)=215+130=430+130=530=16P(L) = P(B \cap L) + P(G \cap L) = \frac{2}{15} + \frac{1}{30} = \frac{4}{30} + \frac{1}{30} = \frac{5}{30} = \boxed{\frac{1}{6}}

Part b(ii)

120\frac{1}{20}

A test tube is broken by a left-handed student

  1. Identify the paths leading to a test tube being broken by a left-handed student.

    These are (Boy and Left-handed and Broken) or (Girl and Left-handed and Broken).

  2. Calculate the probability for each path.

    P(B∩L∩D)=P(B)×P(L∣B)×P(D∣L)=23×15×310=6150=125P(B \cap L \cap D) = P(B) \times P(L|B) \times P(D|L) = \frac{2}{3} \times \frac{1}{5} \times \frac{3}{10} = \frac{6}{150} = \frac{1}{25} P(G∩L∩D)=P(G)×P(L∣G)×P(D∣L)=13×110×310=3300=1100P(G \cap L \cap D) = P(G) \times P(L|G) \times P(D|L) = \frac{1}{3} \times \frac{1}{10} \times \frac{3}{10} = \frac{3}{300} = \frac{1}{100}

  3. Sum the probabilities of these mutually exclusive paths.

    P(D∩L)=P(B∩L∩D)+P(G∩L∩D)=125+1100=4100+1100=5100=120P(D \cap L) = P(B \cap L \cap D) + P(G \cap L \cap D) = \frac{1}{25} + \frac{1}{100} = \frac{4}{100} + \frac{1}{100} = \frac{5}{100} = \boxed{\frac{1}{20}}

Part b(iii)

112\frac{1}{12}

A test tube is broken by a right-handed student

  1. Identify the paths leading to a test tube being broken by a right-handed student.

    These are (Boy and Right-handed and Broken) or (Girl and Right-handed and Broken).

  2. Calculate the probability for each path.

    P(B∩R∩D)=P(B)×P(R∣B)×P(D∣R)=23×45×110=8150=475P(B \cap R \cap D) = P(B) \times P(R|B) \times P(D|R) = \frac{2}{3} \times \frac{4}{5} \times \frac{1}{10} = \frac{8}{150} = \frac{4}{75} P(G∩R∩D)=P(G)×P(R∣G)×P(D∣R)=13×910×110=9300=3100P(G \cap R \cap D) = P(G) \times P(R|G) \times P(D|R) = \frac{1}{3} \times \frac{9}{10} \times \frac{1}{10} = \frac{9}{300} = \frac{3}{100}

  3. Sum the probabilities of these mutually exclusive paths.

    P(D∩R)=P(B∩R∩D)+P(G∩R∩D)=475+3100=16300+9300=25300=112P(D \cap R) = P(B \cap R \cap D) + P(G \cap R \cap D) = \frac{4}{75} + \frac{3}{100} = \frac{16}{300} + \frac{9}{300} = \frac{25}{300} = \boxed{\frac{1}{12}}

Part b(iv)

1315\frac{13}{15}

A test tube is not broken in any session

  1. The probability that a test tube is not broken, P(D′)P(D'), is 1 minus the probability that a test tube is broken, P(D)P(D).

  2. Calculate the total probability of a test tube being broken, P(D)P(D).

    This is the sum of a test tube being broken by a left-handed student and a test tube being broken by a right-handed student.

    P(D)=P(D∩L)+P(D∩R)P(D) = P(D \cap L) + P(D \cap R)

    Using the results from (b)ii and (b)iii:

    P(D)=120+112P(D) = \frac{1}{20} + \frac{1}{12}

  3. Find a common denominator and sum.

    P(D)=360+560=860=215P(D) = \frac{3}{60} + \frac{5}{60} = \frac{8}{60} = \frac{2}{15}

  4. Calculate the probability that a test tube is not broken.

    P(D′)=1−P(D)=1−215=1315P(D') = 1 - P(D) = 1 - \frac{2}{15} = \boxed{\frac{13}{15}}

    2 zimekamilika. Zimebaki 3 leo — tuma inayofuata.

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