Probability Test 11

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Probability Test 11

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Weekly Quiz Competition

Question 1
1 / -0

Box $$I$$ contains $$5$$ red and $$4$$ blue balls, while box $$II$$ contains $$4$$ red and $$2$$ blue balls. A fair die is thrown. If it turns up a multiple of $$3$$, a ball is drawn from the box $$I$$ else a ball is drawn from box $$II$$. Find the probability of the event ball drawn is from the box $$I$$ if it is blue.

A
$$\displaystyle \frac{1}{6}$$

B
$$\displaystyle \frac{15}{19}$$

C
$$\displaystyle \frac{4}{19}$$

D
$$\displaystyle \frac{10}{27}$$

Solution

Box $$I$$ contains $$:5$$ red $$+{4}$$ blue
Box $$II$$ contains $$:4$$ red $$+{2}$$ blue
A ball is taken from Box $$I$$ if a multiple of $$3$$ comes up i.e, $$3$$ and $$6.$$

Ball is taken from Box $$II$$ when $$1,2,4$$ and $$5$$ turns up.

$$\Rightarrow$$ Event of picking up from Box $$I=P(A_1)=\dfrac{2}{6}=\dfrac{1}{3}.$$

$$\Rightarrow$$ Event of picking up from Box $$II=P(A_2)=\dfrac{4}{6}=\dfrac{2}{3}.$$

$$\Rightarrow R=$$ event of drawing a blue ball

$$=P(A_1)P(\dfrac R{A_1})+P(A_2)P(\dfrac{R}A_2)$$

$$=\dfrac{1}{3}\times\dfrac{4}{9}+\dfrac{2}{3}\times\dfrac{2}{6}$$

$$=\dfrac{4}{27}+\dfrac{4}{18}$$

$$=\dfrac{10}{27}.$$
Hence, the answer is $$\dfrac{10}{27}.$$
Question 2
1 / -0

A school has five houses A, B, C, D and E. A class has 23 students, 4 from house A, 8. from house B, 5 from house C, 2 from house 0 and rest from house E. A single student is selected at random ,to be the class monitor. The probability that the selected student is not from A, Band C is?

A
$$\displaystyle \frac{4}{23}$$

B
$$\displaystyle \frac{6}{23}$$

C
$$\displaystyle \frac{8}{23}$$

D
$$\displaystyle \frac{17}{23}$$

Solution

Total number of students, n(S) = 23

Number of students in houses A,B and C

= 4+8+5 = 17

∴ Remaining students = 23 - 17 = 6 n(E) = 6

So, probability that the selected students is not from A,B and C

$$P(E)=\dfrac{6}{23}$$
Question 3
1 / -0

There are three different Urns, Urn-I, Urn-II and Urn-III containing 1 Blue, 2 Green, 2 Blue, 1 Green, 3 Blue, 3 Green balls respectively. If two Urns are randomly selected and a ball is drawn from each Urn and if the drawn balls are of different colours then the probability that chosen Urn was Urn-I and Urn-II is

A
$$\dfrac {1}{7}$$

B
$$\dfrac {5}{13}$$

C
$$\dfrac {5}{14}$$

D
$$\dfrac {5}{7}$$

Solution

Required probability$$\displaystyle =\dfrac {\dfrac {1}{3}\left (\dfrac {1}{3}.\dfrac {1}{3}+\dfrac {2}{3}.\dfrac {2}{3}\right )}{\dfrac {1}{3}\left (\dfrac {1}{3}.\dfrac {2}{3}.\dfrac {2}{3}\right )+\dfrac {1}{3}\left (\dfrac {2}{3}.\dfrac {3}{6}+\dfrac {1}{3}.\dfrac {3}{6}\right )+\dfrac {1}{3}\left (\dfrac {3}{6}.\dfrac {2}{3}+\dfrac {3}{6}.\dfrac {1}{3}\right )}$$
$$\displaystyle =\dfrac {\dfrac {5}{9}}{\dfrac {5}{9}+\dfrac {9}{18}+\dfrac {9}{18}}\\ =\dfrac {5}{14}$$
Question 4
1 / -0

Two aeroplanes $$I$$ and $$II$$ bomb a target in succession. The probability of $$I$$ and $$II$$ scoring a hit correctly are $$0.3$$ and $$0.2$$ respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is

A
$$0.06$$

B
$$0.14$$

C
$$0.32$$

D
$$0.7$$

Solution

$$\mathbf{{\text{Step -1: Stating all the possibilities}}{\text{.}}}$$
$${\text{Writing down all the possibilities that the target is hit by the second plane}}{\text{.}}$$
$${\text{First plane fails and then the second one succeed}}{\text{.}}$$
$${\text{The first one fails, then second fails, the again first one fails, and then the second one succeed}}{\text{.}}$$
$$\text{Similarly, we can observe that all cases in which the second plane succeed }$$
$${\text{at last will be considered}}{\text{.}}$$
$$\mathbf{{\text{Step -2: Finding the probability}}{\text{.}}}$$
$${\text{We know that the probability of first plane hitting correctly is 0}}{\text{.3}}$$
$${\text{and that of the second plane is 0}}{\text{.2}}{\text{.}}$$
$${\text{So let A be the event when first plane succeeds and B be the event when second plane succeeds}}{\text{.}}$$
$${\text{P(A) = 0}}{\text{.3 and P(B) = 0}}{\text{.2}}$$
$${\text{Let E be the event that the target is hit by the second plane}}{\text{.So we can write,}}$$
$${\text{P(E) = P(1 - A)P(B) + P(1 - A)P(1 - B)P(1 - A)P(B) + }}$$
$${\text{P(1 - A)P(1 - B)P(1 - A)P(1 - A)P(1 - B)P(B) + }}......\infty $$
$${\text{The expression written above represent that first fails and second succeeded, }}$$
$${\text{first fails, second, fails, again first fails and then second succeeded,}}....$$
$${\text{So P(E) = 0}}{\text{.7}} \times 0.2 + 0.7 \times 0.8 \times 0.7 \times 0.2 + 0.7 \times 0.8 \times 0.7 \times 0.8 \times 0.7 \times 0.2 + .....\infty $$
$${\text{P(E) = (0}}{\text{.7}} \times 0.2)(1 + 0.8 \times 0.7 + 0.8 \times 0.7 \times 0.8 \times 0.7 + .....\infty )$$
$${\text{From this we can observe that }}1 + 0.8 \times 0.7 + 0.8 \times 0.7 \times 0.8 \times 0.7 + .....\infty {\text{ forms a GP with }}$$
$${\text{infinite terms and common ratio(0}}{\text{.56) < 1}}{\text{.}}$$
$${\text{As we know that sum of such a GP = }}\dfrac{{\text{a}}}{{1 - {\text{r}}}}$$
$${\text{P(E) = (0}}{\text{.14)}} \times \dfrac{{\text{1}}}{{1 - 0.56}}$$
$${\text{P(E) = }}\dfrac{{0.14}}{{0.44}}$$$${\text{ = }}\dfrac{7}{{22}}.$$
$$\mathbf{{\text{}}{\text{ Thus, the probability that the target is hit by the second plane is }}\dfrac{7}{{22}}{\text{ = 0}}{\text{.32}}{\text{.}}}$$
Question 5
1 / -0

If $$E_{1}$$ and $$E_{2}$$ are two events such that $$\displaystyle P\left ( E_{1} \right )=\frac{1}{4}, P\left ( E_{2} \right )=\frac{1}{2}; P\left ( \frac{E_{1}}{E_{2}} \right )=\frac{1}{4},$$ then choose the correct options.

A
$$E_{1}$$ and $$E_{2}$$ are mutually exclusive

B
$$E_{1}$$ & $$E_{2}$$ are dependent

C
$$E_{1}$$ and $$E_{2}$$ are independent

D
$$E_{1}$$ and $$E_{2}$$ are exhaustive

Solution

Since $$\displaystyle P\left ( \frac{E_{1}}{E_{2}} \right )=P\left ( E_{1} \right )$$
$$\Rightarrow $$ $$E_{1}$$ and $$E_{2}$$ are independent of each other
Also since $$P(E_{1}\cup E_{2})=P\left ( E_{1} \right )+P\left ( E_{2} \right )-P\left ( E_{1} \right )|P\left ( E_{2} \right )\neq 1$$
Hence events are not exhaustive. Independent events can't be mutually exclusive.
Hence only (i) is correct
Further since $$E_{1}$$ & $$E_{2}$$ are independent; $$E_{1}$$ and $$\bar{E_{2}}$$ or $$\bar{E_{1}},$$ $$E_{2}$$ are $$\bar{E_{1}}, \bar{E_{2}}$$ are also independent.
Hence $$\displaystyle P\left ( \frac{\bar{E_{1}}}{E_{2}} \right )=P\left ( \bar{E_{1}} \right )=\frac{3}{4}$$ and $$\displaystyle P\left ( \frac{E_{2}}{\bar{E_{1}}} \right )=P\left ( E_{2} \right )=\frac{1}{2}$$
Question 6
1 / -0

35

A
212

B
532

C
244

D
733

Solution
Question 7
1 / -0

Which set is shaded in the above diagram?

A
$$A\cap C$$

B
$$A\cap B\cap C$$

C
$$A\cup(B\cap C)$$

D
$$A\cap(B\cup C)$$

Solution

$$=A\cup(B \cap C) $$

Hence, this is the answer.
Question 8
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Probability of solving specific independently by $$A$$ and $$B$$ are $$\displaystyle\frac { 1 }{ 2 } $$ and $$\displaystyle\frac { 1 }{ 3 } $$ respectively. If both try to solve the problem independently, find the probability that
(i) the problem is solved
(ii) exactly one of them solves the problem.

A
$$0.67,0.78$$

B
$$0.66,0.50$$

C
$$0.67,0.98$$

D
$$0.66,0.44$$

Solution

Probability of solving the problem by $$A$$, $$P\left( A \right) = \displaystyle\frac { 1 }{ 2 }$$
Probability of solving the problem by $$B$$, $$P\left( B \right) = \displaystyle\frac { 1 }{ 3 }$$
Since the problem is solved independently by $$A$$ and $$B$$,
$$\therefore P\left( AB \right) = P\left( A \right) \cdot P\left( B \right) = \displaystyle\frac { 1 }{ 2 } \times \displaystyle\frac { 1 }{ 3 } = \displaystyle\frac { 1 }{ 6 }$$
$$ P\left( A\prime \right) = 1 - P\left( A \right) = 1 - \displaystyle\frac { 1 }{ 2 } = \displaystyle\frac { 1 }{ 2 }$$
$$ P\left( B\prime \right) = 1 - P\left( B \right) = 1 - \displaystyle\frac { 1 }{ 3 } = \displaystyle\frac { 2 }{ 3 }$$
(i) Probability that the problem is solved $$ = P\left( A\cup B \right)$$
$$ = P\left( A \right) + P\left( B \right) - P\left( AB \right)$$
$$ = \displaystyle\frac { 1 }{ 2 } + \displaystyle\frac { 1 }{ 3 } - \displaystyle\frac { 1 }{ 6 }$$
$$ = \displaystyle\frac { 4 }{ 6 }$$
$$ = \displaystyle\frac { 2 }{ 3 }=0.66$$
(ii) Probability that exactly one of them solves the problem is given by,
$$ P\left( A \right) \cdot P\left( B\prime \right) + P\left( B \right) \cdot P\left( A\prime \right)$$
$$ = \displaystyle\frac { 1 }{ 2 } \times \displaystyle\frac { 2 }{ 3 } + \displaystyle\frac { 1 }{ 2 } \times \displaystyle\frac { 1 }{ 3 }$$
$$ = \displaystyle\frac { 1 }{ 3 } + \displaystyle\frac { 1 }{ 6 }$$
$$ = \displaystyle\frac { 1 }{ 2 }=0.50 $$
Question 9
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Given that the events $$A$$ and $$B$$ are such that $$P\left( A \right) = \displaystyle\frac { 1 }{ 2 } , P\left( A\cup B \right) = \displaystyle\frac { 3 }{ 5 } $$ and $$P\left( B \right) =p$$. Find $$p$$ if they are (i) mutually exclusive (ii) independent.

A
$$0.5,0.6$$

B
$$0.1,0.2$$

C
$$0,2,0.4$$

D
$$0.1,0.6$$

Solution

It is given that $$P\left( A \right) = \displaystyle\frac { 1 }{ 2 } , P\left( A\cup B \right) = \displaystyle\frac { 3 }{ 5 } $$ and $$P\left( B \right) = p$$
(i) When $$A$$ and $$B$$ are mutually exclusive, $$A\cap B=\phi $$
$$\therefore P\left( A\cap B \right) =0$$
It is known that, $$P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) $$
$$\Rightarrow \displaystyle\frac { 3 }{ 5 } = \displaystyle\frac { 1 }{ 2 } + p - 0$$
$$\Rightarrow p = \displaystyle\frac { 3 }{ 5 } - \displaystyle\frac { 1 }{ 2 } = \displaystyle\frac { 1 }{ 10 }=0.1 $$
(ii) When $$A$$ and $$B$$ are independent, $$P\left( A\cap B \right) =P\left( A \right) \cdot P\left( B \right) = \displaystyle\frac { 1 }{ 2 } p$$
It is known that, $$P\left( A\cup B \right) =P\left( A \right) + P\left( B \right) -P\left( A\cap B \right) $$
$$\Rightarrow \displaystyle\frac { 3 }{ 5 } = \displaystyle\frac { 1 }{ 2 } + p - \displaystyle\frac { 1 }{ 2 } p$$
$$\Rightarrow \displaystyle\frac { 3 }{ 5 } = \displaystyle\frac { 1 }{ 2 } + \displaystyle\frac { p }{ 2 } $$
$$\Rightarrow \displaystyle\frac { p }{ 2 } = \displaystyle\frac { 3 }{ 5 } -\displaystyle\frac { 1 }{ 2 } = \displaystyle\frac { 1 }{ 10 } $$
$$\Rightarrow p = \displaystyle\frac { 2 }{ 10 } = \displaystyle\frac { 1 }{ 5 }=0.2 $$
Question 10
1 / -0

A coin is tossed and a single $$6$$-sided die is rolled. Find the probability of landing on the tail side of the coin and rolling $$4$$ on the die.

A
$$\dfrac{1}{12}$$

B
$$\dfrac{6}{5}$$

C
$$\dfrac{4}{3}$$

D
$$\dfrac{3}{4}$$

Solution

$$P$$ (tail) $$=$$ $$\dfrac{1}{2}$$ and $$P(4) =$$ $$\dfrac{1}{6}$$

$$P$$ (tail and $$4$$) $$=$$ $$P$$(tail) $$. P(4)$$
$$=$$$$\cfrac{1}{2}\times \cfrac{1}{6}$$ $$=$$ $$\cfrac{1}{12}$$

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Re-Attempt Weekly Quiz Competition

Probability Test 11

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Probability Test 11

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SHARING IS CARING

Question 1

$$\displaystyle \frac{1}{6}$$

$$\displaystyle \frac{15}{19}$$

$$\displaystyle \frac{4}{19}$$

$$\displaystyle \frac{10}{27}$$

Solution

Question 2

$$\displaystyle \frac{4}{23}$$

$$\displaystyle \frac{6}{23}$$

$$\displaystyle \frac{8}{23}$$

$$\displaystyle \frac{17}{23}$$

Solution

Question 3

$$\dfrac {1}{7}$$

$$\dfrac {5}{13}$$

$$\dfrac {5}{14}$$

$$\dfrac {5}{7}$$

Solution

Question 4

$$0.06$$

$$0.14$$

$$0.32$$

$$0.7$$

Solution

Question 5

$$E_{1}$$ and $$E_{2}$$ are mutually exclusive

$$E_{1}$$ & $$E_{2}$$ are dependent

$$E_{1}$$ and $$E_{2}$$ are independent

$$E_{1}$$ and $$E_{2}$$ are exhaustive

Solution

Question 6

212

532

244

733

Solution

Question 7

$$A\cap C$$

$$A\cap B\cap C$$

$$A\cup(B\cap C)$$

$$A\cap(B\cup C)$$

Solution

Question 8

$$0.67,0.78$$

$$0.66,0.50$$

$$0.67,0.98$$

$$0.66,0.44$$

Solution

Question 9

$$0.5,0.6$$

$$0.1,0.2$$

$$0,2,0.4$$

$$0.1,0.6$$

Solution

Question 10

$$\dfrac{1}{12}$$

$$\dfrac{6}{5}$$

$$\dfrac{4}{3}$$

$$\dfrac{3}{4}$$

Solution

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