Maths Solve the Following Question.(2 Marks)

Testing two fuses one-by-one at random, without replacement from 7 can be done in

${ }^7 C _1 \times{ }^6 C _1$ ways.

$\therefore n ( S )={ }^7 C _1 \times{ }^6 C _1=7 \times 6=42$

Let event A: Getting defective fuses in the first two tests without replacement. There are two defective fuses.

$\begin{aligned} & \therefore n ( A )={ }^2 C _1 \times{ }^1 C _1=2 \times 1=2 \\ & \therefore P ( A )=\frac{n(A)}{n(S)}=\frac{2}{42}=\frac{1}{21}\end{aligned}$

B: Event of drawing a green ball and placing a red ball

C: Event of drawing a red ball in the second draw

$\begin{aligned} & P(A)=\frac{5}{7} \\ & P(B)=\frac{2}{7} \\ & P(C / A)=\frac{4}{7} \\ & P(C / B)=\frac{6}{7}\end{aligned}$

Required probability

$\begin{aligned} & P(C)=P(A) P(C / A)+P(B) P(C / B) \\ & =\frac{5}{7} \times \frac{4}{7}+\frac{2}{7} \times \frac{6}{7} \\ & =\frac{32}{49}\end{aligned}$

turning up is proportional to 1, 2, 3, 4, 5, 6.

Let k be the common ration of proportionality.

∴ The probabilities of the faces with 1, 2, 3, 4, 5, 6 dots turning up are 1k , 2k, 3k, 4k, 5k, 6k respectively.

Since sum of the probabilities = 1,

k(1 + 2+ ….. + 6) = 1

$\begin{aligned} & k \left(\frac{6 \times 7}{2}\right)=1 \\ & k =\frac{1}{21}\end{aligned}$

Required probability = P(1) + P(3) + P(5)

$\begin{aligned} & =\frac{1}{21}+\frac{3}{21}+\frac{5}{21} \\ & =\frac{9}{21} \\ & =\frac{3}{7}\end{aligned}$

$
\therefore \mathrm{P}(\mathrm{B} / \mathrm{A})=\frac{12}{52}=\frac{3}{13}=\mathrm{P}(\mathrm{B}) \text {. }
$

Since first black ball is not replaced in the urn, therefore now we have 9 balls containing 3 black balls.
$\therefore$ Probability of getting second black ball under the condition that first black is not replaced
$
\begin{aligned}
& \text { in the pack }=P(B / A)=\frac{3}{9} \\
& \therefore P(\text { both are black balls })=P(A \cap B) \\
& =P(A) P(B / A)=\frac{4}{10} \times \frac{3}{9}=\frac{2}{15}
\end{aligned}
$

S = {1, 2, 3, 4, 5, 6}

∴ n(S) = 6 Let event A: 3 or 4 turns up.

∴ A = {3, 4}

∴ n(A) = 2

$\therefore P ( A )=\frac{n(n)}{n(S)}=\frac{2}{6}=\frac{1}{3}$

$P\left(A^{\prime}\right)=1-P(A)=1-\frac{1}{3}=\frac{2}{3}$

∴ Odds against the event A are P(A’) : P(A)

$\begin{aligned} & =\frac{2}{3}: \frac{1}{3} \\ & =2: 1\end{aligned}$

event $A _2$ : Disease in mild form,

event $A_3$ : Subject does not have disease,

event B: Subject tests positive.

$P\left(A_1\right)=0.02, P\left(A_2\right)=0.1, P\left(A_3\right)=0.88$

The probability of testing positive is $\frac{9}{10}$ if the subject has the serious form, $\frac{6}{10}$ if the

subject has the mild form, and $\frac{1}{10}$ if the subject doesn't have the disease

$\begin{aligned} & \therefore P\left(B / A_1\right)=0.9, P\left(B / A_2\right)=0.6, P\left(B / A_3\right)=0.1 \\ & P(B)=P\left(A_1\right) P\left(B / A_1\right)+P\left(A_2\right) P\left(B / A_2\right)+P\left(A_3\right) P\left(B / A_3\right) \\ & =0.02 \times 0.9+0.1 \times 0.6+0.88 \times 0.1 \\ & =0.166\end{aligned}$

Required probability = P(A1/B) By Baye’s theorem

$\begin{aligned} P \left( A _1 / B \right) & =\frac{ P \left( A _1\right) P \left( B / A _1\right)}{ P ( B )} \\ & =\frac{0.9 \times 0.02}{0.166} \\ & =0.108\end{aligned}$

$\therefore P\left(B_i\right)=\frac{1}{3}$

If Bag 1 is chosen, it has 75 red and 25 blue marbles.

∴ Probability that the chosen marble is red under the condition that it is from Bag 1 = P(R/B1)

$\begin{aligned} & =\frac{{ }^{75} C _1}{{ }^{100} C _1} \\ & =\frac{75}{100} \\ & =0.75\end{aligned}$

Similarly we get,

$\begin{aligned} & P\left(R / B_2\right)=\frac{60}{100}=0.60 \\ & P\left(R / B_3\right)=\frac{45}{100}=0.45 \\ & \therefore \text { Required probability } \\ & P(R)=P\left(B_1\right) P\left(R / B_1\right)+P\left(B_2\right) P\left(R / B_2\right)+P\left(B_3\right) P\left(R / B_3\right) \\ & =\frac{1}{3}(0.75)+\frac{1}{3}(0.60)+\frac{1}{3}(0.45) \\ & =\frac{1}{3}(1.8) \\ & =0.60\end{aligned}$

∴ Sample space S = {BB, BG, GB, GG}

∴ n(S) = 4 Let event A: At least one of the children is a girl.

∴ A = {GG, GB, BG}

∴ n(A) = 3

$\therefore P ( A )=\frac{n(A)}{n(S)}=\frac{3}{4}$

Let event B: Both children are girls.

∴ B = {GG}

∴ n(B) = 1

$\therefore P ( B )=\frac{n(B)}{n(S)}=\frac{1}{4}$

Also, A ∩ B = B

$\therefore P(A \cap B)=P(B)=\frac{1}{4}$

∴ Required probability = P(B/A)

$\begin{aligned} & =\frac{P(B \cap A)}{P(A)} \\ & =\frac{\frac{1}{4}}{\frac{3}{4}} \\ & =\frac{1}{3}\end{aligned}$

$P(A)=\frac{5}{8}$

After drawing the first ball, without replacing it into the bag a second ball is drawn from the remaining 7 balls.

$\begin{aligned} & \therefore P(B / A)=\frac{4}{7} \\ & \therefore P(B o t h \text { balls are white })=P(A \cap B) \\ & =P(A) \cdot P(B / A) \\ & =\frac{5}{8} \times \frac{4}{7} \\ & =\frac{5}{14}\end{aligned}$

$\therefore P(A)=\frac{{ }^{12} C_1}{{ }^{52} C_1}=\frac{12}{52}=\frac{3}{13}$

Let event B: The second card drawn is a face card. Since the first card is not replaced in the pack, we now have 51 cards, out of which 11 are face cards.

∴ Probability that the second card is a face card under the condition that the first card is

not replaced in the pack $= P ( B / A )=\frac{{ }^{11} C_1}{{ }^{51} C_1}=\frac{11}{51}$

∴ Required probability = P(A ∩ B) = P(B/A) . P(A)

$\begin{aligned} & =\frac{11}{51} \times \frac{3}{13} \\ & =\frac{11}{221}\end{aligned}$

Two balls are drawn from 15 balls without replacement.

$\therefore n ( S )={ }^{15} C _1 \times{ }^{14} C _1=15 \times 14=210$

Let event A: At least one ball is black.

i.e., the first ball is black, and the second ball is non-black or the first ball is non-black and the second ball is black,

or both the first and second balls are black.

$\begin{aligned} & \therefore n ( A )={ }^4 C _1 \times{ }^{11} C _1+{ }^{11} C _1 \times{ }^4 C _1+{ }^4 C _1 \times{ }^3 C _1 \\ & =4 \times 11+11 \times 4+4 \times 3 \\ & =100 \\ & \therefore P ( A )=\frac{n(A)}{n(S)}=\frac{100}{210}=\frac{10}{21}\end{aligned}$

Check:

Required probability = 1 – P(no black ball in two balls)

$=1-\frac{{ }^{11} C_2}{{ }^{15} C_2}=1-\frac{11 \times 10}{15 \times 14}=1-\frac{11}{21}=\frac{10}{21}$

$\begin{aligned} & \therefore P(A)=\frac{80}{100}=\frac{4}{5} \\ & \text { and } P(B)=\frac{60}{100}=\frac{3}{5} \\ & P\left(A^{\prime}\right)=1-P(A)=1-\frac{4}{5}=\frac{1}{5} \\ & \text { and } P\left(B^{\prime}\right)=1-P(B)=1-\frac{3}{5}=\frac{2}{5}\end{aligned}$

∴ P(A and B contradict each other) = P(A speaks the truth and B lies) + P (A lies and B speaks the truth)

= P(A ∩ B’) + P(A’ ∩ B)

= P(A) P(B’) + P(A’) P(B)

$\begin{aligned} & =\left(\frac{4}{5} \times \frac{2}{5}\right)+\left(\frac{1}{5} \times \frac{3}{5}\right) \\ & =\frac{8}{25}+\frac{3}{25} \\ & =\frac{11}{25}\end{aligned}$

$\therefore P(A)=\frac{{ }^{13} C _1}{{ }^{52} C _1}=\frac{13}{52}=\frac{1}{4}$

(i) Let event B: The second card drawn is a diamond card.

Since the first diamond card is kept aside, we now have 51 cards, out of which 12 are diamond cards.

Probability that the second card is a diamond card under the condition that the first

diamond card is kept aside in the pack $=P(B / A)=\frac{{ }^{12} C _1}{{ }^{51} C _1}=\frac{12}{51}=\frac{4}{17}$

∴ Required probability

= P(A ∩ B) = P(B/A) . P(A)

$\begin{aligned} & =\frac{1}{4} \times \frac{4}{17} \\ & =\frac{1}{17}\end{aligned}$

Let event B: The second card drawn is a diamond card.

Since the first diamond card is replaced in the pack, we now again have 52 cards, out of which 13 are diamond cards.

∴ Probability that the second card is a diamond card under the condition that the first

diamond card is replaced in the pack $= P ( B / A )=\frac{{ }^{13} C _1}{5^2 C _1}=\frac{13}{52}=\frac{1}{4}$

Required probability

= P(A ∩ B) = P(B/A) . P(A)

$\begin{aligned} & =\frac{1}{4} \times \frac{1}{4} \\ & =\frac{1}{16}\end{aligned}$

P(B) = 1 – P(B’)

$\begin{aligned} & =1-\frac{1}{3} \\ & =\frac{2}{3}\end{aligned}$

Since P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

$\begin{aligned} & \frac{5}{6}=P(A)+\frac{2}{3}-\frac{1}{3} \\ & \therefore \frac{5}{6}=P(A)+\frac{1}{3} \\ & \therefore P(A)=\frac{5}{6}-\frac{1}{3}=\frac{1}{2}\end{aligned}$

16 persons in the conference.

These 16 persons can be arranged among themselves around a round table in (16 – 1)! = 15! ways.

∴ n(S) = 15!

Let event A: Two particular professors be seated on either side of the Vice-chancellor.

Those two particular persons sit on either side of a Vice chancellor in ${ }^2 P _2=2$ ! ways.

Thus the remaining 13 persons can be arranged in ${ }^{13} P_{13}=13$ ! ways.

$\begin{aligned} & \therefore n(A)=13 ! 2 ! \\ & \therefore P(A)=\frac{n(A)}{n(S)}=\frac{13 ! \times 2 !}{15 !}=\frac{13 ! \times 2 \times 1}{15 \times 14 \times 13 !}=\frac{1}{105}\end{aligned}$

Since any number of letters can be posted in all 4 post boxes,

so each letter can be posted in different ways.

∴ n(S) = 4 × 4 × 4 × 4

Let event A: Each box contains only one letter.

∴ 1st letter can be posted in 4 different ways.

Since each box contains only one letter, 2nd letter can be posted in 3 different ways.

Similarly, 3rd and 4th letters can be posted in 2 different ways and 1 way respectively.

$\begin{aligned} & \therefore n ( A )=4 \times 3 \times 2 \times 1 \\ & \therefore P ( A )=\frac{n(A)}{n(S)}=\frac{4 \times 3 \times 2 \times 1}{4 \times 4 \times 4 \times 4}=\frac{3}{32}\end{aligned}$

It has 52 complete weeks and two more days.

These two days can be {(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thur), (Thur, Fri), (Fri, Sat), (Sat, Sun)}.

∴ n(S) = 7

Let event E : There are 53 Sundays.

∴ E = {(Sun, Mon), (Sat, Sun)}

∴ n(E) = 2

$\therefore P ( E )=\frac{n(E)}{n(S)}=\frac{2}{7}$