Solve the Following Question.(4 Marks)

Question 14 Marks

The odds against student $X$ solving a business statistics problem are $8: 6$ and the odds in favour of student $Y$ solving the same problem are $14: 16$:
1.What is the chance that the problem will be solved, if they try independently?
2.What is the probability that neither solves the problem?

Answer

1.t $A$ be the event that $X$ solves the problem $B$ be the event that $Y$ solves the problem. Since the odds against student $X$ solving the problem are $8: 6$
$\therefore$ Probability of occurrence of event $A$ is given by
$
\begin{aligned}
& P(A)=\frac{6}{8+6}=\frac{6}{14} \\
& \text { and } P\left(A^{\prime}\right)=1-P(A) \\
& =1-\frac{6}{14} \\
& =\frac{8}{14}
\end{aligned}
$
Also, the odds in favour of student $Y$ solving the problem are 14:16
$\therefore$ Probability of occurrence of event $B$ is given by
$
\begin{aligned}
& P(B)=\frac{14}{14+16}=\frac{14}{30} \text { and } \\
& P\left(B^{\prime}\right)=1-P(B) \\
& =1-\frac{14}{30} \\
& =\frac{16}{30}
\end{aligned}
$
Now $A$ and $B$ are independent events.
$\therefore A^{\prime}$ and $B^{\prime}$ are independent events.
$\therefore A^{\prime} \cap B^r=$ Event that neither solves the problem
$=P\left(A^{\prime} \cap B^{\prime}\right)$
$=P\left(A^{\prime}\right) \cdot P\left(B^{\prime}\right)$
$=\frac{8}{14} \times \frac{16}{30}$
$=\frac{32}{105}$
$A \cup B=$ the event that the problem is solved
$\therefore \mathrm{P}($ problem will be solved $)=\mathrm{P}(\mathrm{A} \cup \mathrm{B})$
$=1-P(A \cup B)^{\prime}$
$=1-P\left(A^{\prime} \cap B^{\prime}\right)$
$=1-\frac{32}{105}$
$=1-\frac{73}{105}$

2.P$ (neither solves the problem) $=P\left(A^{\prime} \cap B^{\prime}\right)$
$=P\left(A^{\prime}\right) P\left(B^{\prime}\right)$
$=\frac{8}{14} \times \frac{16}{30}$
$=\frac{32}{105}$

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Question 24 Marks

In a group of students, there are 3 boys and 4 girls. Four students are to be selected at random from the group. Find the probability that either 3 boys and 1 girl or 3 girls and 1 boy are selected.

Answer

The group consists of 3 boys and 4 girls i.e., 7 students.
4 students can be selected from this group in ${ }^7 \mathrm{C}_4$
$
\begin{aligned}
& =\frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} \\
& =35 \text { ways. } \\
& \therefore \mathrm{n}(\mathrm{S})=35
\end{aligned}
$
Let $\mathrm{A}$ be the event that 3 boys and $1 \mathrm{girl}$ are selected.
3 boys can be selected in ${ }^3 C_3$ ways while a girl can be selected in ${ }^4 C_1$ ways.
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{A})={ }^3 \mathrm{C}_3 \times{ }^4 \mathrm{C}_1=4 \\
& \therefore \mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{4}{35}
\end{aligned}
$
Let $B$ be the event that 3 girls and 1 boy are selected.
3 girls can be selected in ${ }^4 \mathrm{C} 3$ ways and a boy can be selected in ${ }^3 \mathrm{C}_1$ ways.
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{B})={ }^4 \mathrm{C}_3 \times{ }^3 \mathrm{C}_1=12 \\
& \therefore \mathrm{P}(\mathrm{B})=\frac{n(B)}{n(S)}=\frac{12}{35}
\end{aligned}
$
Since $A$ and $B$ are mutually exclusive and exhaustive events
$
\begin{aligned}
& \therefore P(A \cap B)=0 \\
& \therefore \text { Required probability }=P(A \cup B) \\
& =P(A)+P(B) \\
& =\frac{4}{35}+\frac{12}{35} \\
& =\frac{16}{35}
\end{aligned}
$

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Question 34 Marks

A card is drawn from a well-shuffled pack of 52 cards. Find the probability of it being a heart or a queen.

Answer

One card can be drawn from the pack of 52 cards in ${ }^{52} \mathrm{C}_1=52$ ways
$
\therefore \mathrm{n}(\mathrm{S})=52
$
Also, the pack of 52 cards consists of 13 heart cards and 4 queen cards
Let $\mathrm{A}$ be the event that a card drawn is the heart.
A heart card can be drawn from 13 heart cards in ${ }^{13} C_1$ ways
$
\begin{aligned}
& \therefore \mathrm{n}(\mathrm{A})={ }^{13} \mathrm{C}_1 \\
& \therefore \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_1}{{ }^{52}}=\frac{13}{52}
\end{aligned}
$
Let $B$ be the event that a card drawn is queen.
A queen card can be drawn from 4 queen cards in ${ }^4 C_1$ ways
$
\begin{aligned}
& \therefore \mathrm{n}(B)={ }^4 \mathrm{C}_1 \\
& \therefore \mathrm{P}(B)=\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^4 \mathrm{C}_1}{52}=\frac{4}{52}
\end{aligned}
$
There is one queen card out of 4 which is also a heart card
$
\begin{aligned}
& \therefore n(A \cap B)={ }^1 C_1 \\
& \therefore P(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{{ }^1 C_1}{52}=\frac{1}{52} \\
& \therefore P(\text { card is a heart or a queen })=P(A \cup B) \\
& =P(A)+P(B)-P(A \cap B) \\
& =\frac{13}{52}+\frac{4}{52}-\frac{1}{52} \\
& =\frac{13+4-1}{52} \\
& =\frac{16}{52} \\
& \therefore P(A \cup B)=\frac{4}{13}
\end{aligned}
$

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Question 44 Marks

Two dice are thrown together. What is the probability that sum of the numbers on two dice is 5 or the number on the second die is greater than or equal to the number on the first die?

Answer

When two dice are thrown, the sample space is
$
S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,
$
$3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$,
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$
$
\therefore \mathrm{n}(\mathrm{S})=36
$
Let $\mathrm{A}$ be the event that sum of numbers on two dice is 5 .
$
\begin{aligned}
& \therefore \mathrm{A}=\{(1,4),(2,3),(3,2),(4,1)\} \\
& \therefore \mathrm{n}(\mathrm{A})=4 \\
& \therefore \mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{4}{36}
\end{aligned}
$
Let $B$ be the event that number on second die is greater than or equal to number on first die.
$
\begin{aligned}
& B=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(2,3),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3, \\
& 6),(4,4),(4,5),(4,6),(5,5),(5,6),(6,6)\} \\
& \therefore \mathrm{n}(B)=21 \\
& \therefore P(B)=\frac{n(B)}{n(S)}=\frac{21}{36} \\
& \text { Now, } A \cap B=\{(1,4),(2,3)\} \\
& \therefore n(A \cap B)=2 \\
& \therefore P(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{2}{36} \\
& \therefore \text { Required probability }=P(A \cup B) \\
& P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
& =\frac{4}{36}+\frac{21}{36}-\frac{2}{36} \\
& =\frac{23}{36}
\end{aligned}
$

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Maths (commerce) Solve the Following Question.(4 Marks)

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