5 Marks Questions

Question 15 Marks

One by one 3 cards are drawn from a well-shuffled deck of 52 cards without replacement. Find the probability of first two cards are Ace and third card is King.

Answer

Let event A : Getting card is Ace and K : Getting card is King. Clearly, we have to find out $P ( AAK )$. Now $\quad
P(A)=\frac{4}{52}=\frac{1}{13}
$
To find $P(A \mid A)$, first card is Ace then probability of getting second card is Ace. Now in pack, there is 51 cards in which 3 Aces. So,
$
P(A \mid A)=\frac{3}{51}=\frac{1}{17}
$
Hence $P(K \mid A A)$ is conditional probability of third card is King when first two cards were Aces. Now in pack 50 cards will remain.
$
\therefore \quad P(AKK)=\frac{4}{50}=\frac{2}{25}
$
By multiplication rule of probability, we find that
$
\begin{aligned}
P(AAK) & =P(A) \cdot P(A \mid A) \cdot P(K \mid AA) \\
& =\frac{1}{13} \times \frac{1}{17} \times \frac{2}{25} \\
& =\frac{2}{5525} \quad \text { }
\end{aligned}
$

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Question 25 Marks

A dice rolled two times and sum of appeared number found 7 . Find the conditional probability of getting 3 at least one time.

Answer

Let event E 'Number 3 appear at least one time' and event $F$ 'Sum of appeared number is 7 on both dice' then $\quad E =\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$,
$\quad \quad \quad\quad\quad\quad\quad \quad(1,3),(2,3),(4,3),(5,3),(6,3)\}$ $\text{and}\quad \quad \quad F =\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}$
We know that
$
P(E)=\frac{11}{36}, P(F)=\frac{6}{36}
$

$\text{and} \quad
E \cap F=\{(3,4),(4,3)\}
$
Now $\quad P ( E \cap F )=\frac{2}{36}$
Hence such probability.
$\begin{aligned} P ( E / F ) & =\frac{ P ( E \cap F )}{ P ( F )}=\frac{2 / 36}{6 / 36} \\ & =\frac{2}{6}=\frac{1}{3}\end{aligned}$

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Question 35 Marks

Suppose a girl throw a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether a 'head' or 'tail' is obtained exactly one 'tail', what is the probability that she threw 3, 4, 5 or 6 with the die?

Answer

A dice is rolled we get 6 (1, 2, 3, 4, 5, 6) results.
Let event E₁: getting 1 or 2
E2: getting 3, 4, 5, 6 E : tails appear when coin tossed
$P \left( E _1\right): P$ (getting 1,2 when dice rolled)
$
=\frac{2}{6}=\frac{1}{3}
$
$P \left( E _2\right): P$ (getting 3, 4, 5, 6 when dice rolled)
$
=\frac{4}{6}=\frac{2}{3}
$
Total results when 3 dice rolled [TTT, TTH, THT, HTT, HHT, HTH, THH, HHH\} $=8$.
Methods of getting one tail THH, HTH, HHT P(E|E $\left.E _1\right)$ (getting 1,2 when dice rolled and getting 1 tail tossed 3
$
\text { coins })=\frac{3}{8}
$
Probability of getting tail when one coin tossed $=\frac{1}{2}$ or $P \left( E \mid E _1\right)= P$ (getting 3, 4, 5, 6 when dice rolled and getting tail tossed 1 coin) $=\frac{1}{2}$
By Bayes' theorem,
$
\begin{aligned}
P\left(E_2 \mid E\right) & =\frac{P\left(E_2\right) P\left(E \mid E_2\right)}{P\left(E_2\right) P\left(E \mid E_2\right)+P\left(E_1\right) P\left(E \mid E_1\right)} \\
& =\frac{\frac{2}{3} \times \frac{1}{2}}{\frac{2}{3} \times \frac{1}{2}+\frac{1}{3} \times \frac{3}{8}}=\frac{1 / 3}{\frac{1}{3}+\frac{1}{8}} \\
& =\frac{1 / 3}{11 / 24}=\frac{1}{3} \times \frac{24}{11}=\frac{8}{11} \quad \text { Ans. }
\end{aligned}
$

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Question 45 Marks

A bag has 4 red and 6 black balls and other bag has 3 red and 5 black balls. One of the two bags, is selected at random and a ball is drawn at random. The ball drawn is found to be red. What is the probability that ball comes out from second bag?

Answer

$E _1$ : Choose bag I
$E _2$ : Choose bag II
A : Red ball drawn
$
\begin{aligned}
then \quad P\left(E_1\right) & =P\left(E_2\right)=\frac{1}{2} \\
P\left(A \mid E_1\right) & =P(\text { Red ball drawn from first bag) } \\
& =\frac{4}{10}=\frac{2}{5}
\end{aligned}
$
and $P\left(A \mid E_2\right)=P($ Red ball drawn from second bag)
$
=\frac{3}{8}
$
Probability of ball comes out from second bag where it is a red ball $= P \left( E _2 \mid A \right)$
By Bayes' theorem,
$
\begin{aligned}
P\left(E_2 \mid A\right) & =\frac{P\left(E_2\right) P\left(A \mid E_2\right)}{P\left(E_1\right) P\left(A \mid E_1\right)+P\left(E_2\right) \cdot P\left(A \mid E_2\right)} \\
& =\frac{\frac{1}{2} \times \frac{3}{8}}{\frac{1}{2} \times \frac{2}{5}+\frac{1}{2} \times \frac{3}{8}}=\frac{\frac{3}{8}}{\frac{2}{5}+\frac{3}{8}} \\
& =\frac{\frac{3}{8}}{\frac{31}{40}}=\frac{3}{8} \times \frac{40}{31}=\frac{15}{31}
\end{aligned}
$

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Question 55 Marks

Coloured balls are distributed in $3$ bags as shown in the following table :

Colour of balls
Bag	Black	White	Red
$I$	$1$	$2$	$3$
$1$	$2$	$4$	$1$
$III$	$4$	$5$	$3$

A bag is selected at random and then two balls are randomly drawn from the selected bag. The colours of the balls are black and red. What is the probability that ball drawn is from bag I?

Answer

Let $E_1, E_2, E_3$ be the event of choosing the bag $I, II,$ and $III.$
$\therefore P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}$
Let event E be drawn red and black balls So,
$P\left(E \mid E_1\right) =\frac{{ }^1 C_1 \cdot{ }^3 C_1}{{ }^6 C_2}$
$ =\frac{3}{15}=\frac{1}{5}$
$P\left(E \mid E_2\right) =\frac{{ }^2 C_1 \cdot{ }^1 C_1}{{ }^7 C_2}=\frac{2}{21}$
$P\left(E \mid E_3\right) =\frac{{ }^4 C_1 \cdot{ }^3 C_1}{{ }^{12} C_2}=\frac{4 \times 3}{66}=\frac{12}{66}=\frac{2}{11}$
By Bayes' Theorem,
Such probability
$P\left(E_1 \mid E\right)= \frac{P\left(E_1\right) \cdot P\left(E \mid E_1\right)}{P\left(E_1\right) \cdot P\left(E \mid E_1\right)+P\left(E_2\right) \cdot P\left(E \mid E_2\right)}$
$+P\left(E_3\right) \cdot P\left(E \mid E_3\right)$
$= \frac{\frac{1}{3} \times \frac{1}{5}}{\frac{1}{3} \times \frac{1}{5}+\frac{1}{3} \times \frac{2}{21}+\frac{1}{3} \times \frac{2}{11}}$
$or =\frac{\frac{1}{3} \times \frac{1}{5}}{\frac{1}{3}\left[\frac{1}{5}+\frac{2}{21}+\frac{2}{11}\right]}=\frac{\frac{1}{5}}{\frac{1}{5}+\frac{2}{21}+\frac{2}{11}}$
$=\frac{\frac{1}{5}}{\frac{231+110+210}{5 \times 21 \times 11}}$
$=\frac{1}{5} \times \frac{5 \times 21 \times 11}{551}=\frac{231}{551}$

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