Probability Test

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Probability Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

An arrangement is selected at random from all possible arrangements of five digits written from the digits $$0,1,2,3,\cdots 9$$ with repetition. The probability that the randomly selected arrangement will have largest number $$'8'$$ given that the smallest number is $$'4'$$ is :

A
$$\dfrac {1253}{6480}$$

B
$$\dfrac {513}{4651}$$

C
$$\dfrac {2881}{6480}$$

D
$$\dfrac {1320}{4651}$$
Question 2
1 / -0

A four-digit number is formed by using the digits 1, 2, 4, 8 and 9 without repitition. If one number is selected from those numbers, then what is the probability that it will be an odd number ?

A
$$\dfrac{1}{5}$$

B
$$\dfrac{2}{5}$$

C
$$\dfrac{3}{5}$$

D
$$\dfrac{4}{5}$$

Solution

Total number of outcomes = $$5\times 4\times 3\times 2=120$$
The number of favourable cases = $$2(4\times 3\times2)-=48$$ (i.e., odd numbers )
Therefore,
Required probability = $$\dfrac{48}{120}=\dfrac{2}{5}$$
Question 3
1 / -0

A bouquet from $$11$$ different flowers is to be made so that it contains not less then three flowers . Then the number of the different ways of selecting flowers to from the bouquet

A
$$1972$$

B
$$1952$$

C
$$1981$$

D
$$1947$$

Solution

No. of ways $$=^{11}C_3+^{11}C_4+^{11}C_5+^{11}C_6+^{11}C_7+^{11}C_8+^{11}C_9+^{11}C_{10}+^{11}C_{11}$$
$$\Rightarrow 165+330+462+462+330+165+55+11+1$$
$$\Rightarrow 1981$$
Question 4
1 / -0

$$10$$ persons are seated around a round table. What is the probability that $$4$$ particular persons are always seated together?

A
$$\dfrac 1{20}$$

B
$$\dfrac 4{10}$$

C
$$\dfrac 1{21}$$

D
$$\dfrac 3{20}$$

Solution

10 persons can sit around a table in $$9!$$ ways.
Consider the particular four persons as one unit.
Now, the entities are $$6+1=7$$
These 7 entities can be arranged in $$6!$$ ways.
In the entities itself they can be arranged in $$4!$$ ways.
The required number of arrangements = $$6!4!$$
Probability = $$\dfrac{m}{n}=\dfrac{6!4!}{9!}=\dfrac{1}{21}$$
Question 5
1 / -0

A bag contain $$4$$ white and $$2$$ black balls. Two balls are drawn at random. The probability that they are of the same colour is ________.

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

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

C
$$\dfrac{7}{15}$$

D
$$\dfrac{1}{15}$$

Solution

We assume that there are $$4$$ white balls and $$2$$ black balls.
There are $$\binom{6}{2} = 15$$ total possible ways of drawing two balls from these given $$6$$ balls.
We are interested in the event where the two drawn balls are of the same colour.
For this, we note that the number of ways of drawing $$2$$ white balls is $$\binom{4}{2} = 6$$ whereas the number of ways of drawing $$2$$ black balls is $$\binom{2}{2} = 1$$.
So, the probability that the two drawn balls are of the same colour is $$\frac{6+1}{15} = \frac{7}{15}$$.
Question 6
1 / -0

A box contains $$10$$ items, $$3$$ of which are defective. If $$4$$ are selected at random without replacement, what is the probability that at least $$2$$ of the $$4$$ are defective?

A
$$0.009$$

B
$$0.047$$

C
$$0.026$$

D
None of these

Solution

Here at least $$2$$ items are defective out of $$4$$. So the no. of defective items can be $$2$$ or defective can be $$3$$.
Total no. of defective ites $$= 3$$
If $$2$$ items are defective $$= {^{3}{C}_{2}} \times {^{7}{C}_{2}} = 63$$
If $$3$$ items are defective $$= {^{3}{C}_{3}} \times {^{7}{C}_{1}} = 7$$
Probability that atleast $$2$$ out of $$4$$ items are defective $$= \cfrac{63+7}{{^{10}{C}_{4}}} = \cfrac{70}{210} = \cfrac{1}{3}$$
Question 7
1 / -0

If a fair die is rolled $$4$$ times, then what is the probability that there are at least $$2$$ sixes ?

A
$$\dfrac{19}{144}$$

B
$$\dfrac{25}{216}$$

C
$$\dfrac{125}{216}$$

D
$$\dfrac{175}{216}$$

Solution

From given, we have,

$$6^4$$ total permutations, of them $$5^4$$ are without sixes

we get, $$\left ( \dfrac{5}{6} \right )^4$$

$$P(X\geq 2)1-\left ( \dfrac{5}{6} \right )^4-4\times \dfrac{1}{6}\times \left ( \dfrac{5}{6} \right )^3$$

$$=1-\dfrac{5^4}{6^4}-4 \times \dfrac{5^3}{6^4}$$

$$=\dfrac{6^4-5^4-4(5^3)}{6^4}$$

$$=\dfrac{171}{1296}$$

$$=\dfrac{19}{144}$$
Question 8
1 / -0

If $$5$$ of a company's $$10$$ delivery trucks do not meet emission standard and $$3$$ of them are chosen for inspection, then what is the probability that none of the trucks chosen will meet emission standards ?

A
$$\dfrac{1}{8}$$

B
$$\dfrac{3}{8}$$

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

D
$$\dfrac{1}{4}$$

Solution

$$\textbf{Step-1: Apply combination using given information's.}$$
$$\text{We have,}$$
$$5$$ $$\text {of company's}$$ $$10$$ $$\text{delivery trucks, do not meet emission standard and only}$$ $$3$$
$$\text{trucks selected for inspection.}$$
$$\therefore$$ $$\text{P( none of the 3 truck will pass )}$$ $$=\dfrac{^5C_3}{^{10}C_3}$$
$$=\dfrac{\dfrac{5!}{3!(5-3)!}}{\dfrac{10!}{3!(10-3)!}}$$ $$\boldsymbol{[\because}$$ $$\boldsymbol{{}^nC_r=\dfrac{n!}{r!(n-r)!}]}$$

$$=\dfrac{10}{120}=\dfrac{1}{12}$$
$$\textbf{Hence, option - C is the answer}$$
Question 9
1 / -0

A is targeting to B, B and C are targeting to A.The probability of hitting the target by A, B and Care $$\displaystyle \frac{2}{3}, \displaystyle \frac{1}{2}, \displaystyle \frac{1}{3}$$ respectively. Show that the probability that B hits the target and C does not is $$\displaystyle \frac{1}{2}$$.

A
$$\displaystyle \frac{1}{3}$$

B
$$\displaystyle \frac{1}{2}$$

C
$$\displaystyle \frac{2}{3}$$

D
$$\displaystyle \frac{3}{4}$$

Solution

Probability of A being hit = $$P(E)=1-(Probability\quad of\quad B\quad missing)*(Probability\quad of\quad C\quad missing)=1-(1-\dfrac { 1 }{ 2 } )\times (1-\dfrac { 1 }{ 3 } )=\dfrac { 2 }{ 3 } $$
Probability that A is hit by B but not C = $$P((B\cap \bar { C } )|E)=\dfrac { P(B)*P(\bar { C } ) }{ P(E) } =\dfrac { \dfrac { 1 }{ 2 } *\dfrac { 2 }{ 3 } }{ \dfrac { 2 }{ 3 } } =\dfrac { 1 }{ 2 } $$
Question 10
1 / -0

Two symmetrical dice are thrown $$200$$ times. Getting a sum of $$9$$ points is considered to be a success. The probability distribution of successes is

A
$$\left( 200,\ \cfrac { 1 }{ 9 } ,\ \cfrac { 8 }{ 9 } \right) $$

B
$$\left( 200,\ \cfrac { 2 }{ 9 } ,\ \cfrac { 7 }{ 9 } \right) $$

C
$$\left( 200,\ \cfrac { 4 }{ 9 } ,\ \cfrac { 5 }{ 9 } \right) $$

D
$$\left( 200,\ \cfrac { 1 }{ 4 } ,\ \cfrac { 3 }{ 4 } \right) $$

Solution

Two symmetrical dice are thrown $$200$$ times. $$ n = 200$$
Getting a sum of $$9$$ points is considered to be a success.
Ways of sum $$9$$ are$$ = (3,6),(6,3),(4,5),(5,4) $$ total ways of sum $$9 = 4 $$
Total ways of when $$2$$ die are rolled $$= 36 $$
Probability of success $$ = p =$$ $$ \dfrac {4}{36} = \dfrac {1}{9} $$
So, $$q = (1-p) = ( 1- $$ $$ \dfrac {1}{9} ) = \dfrac {8}{9} $$
The probability distribution of successes is $$= ( n , p , q )$$
$$=\left( 200,\ \dfrac {1}{9}, \dfrac {8}{9} \right)$$

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Probability Test - 17

Result

Probability Test - 17

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SHARING IS CARING

Question 1

$$\dfrac {1253}{6480}$$

$$\dfrac {513}{4651}$$

$$\dfrac {2881}{6480}$$

$$\dfrac {1320}{4651}$$

Question 2

$$\dfrac{1}{5}$$

$$\dfrac{2}{5}$$

$$\dfrac{3}{5}$$

$$\dfrac{4}{5}$$

Solution

Question 3

$$1972$$

$$1952$$

$$1981$$

$$1947$$

Solution

Question 4

$$\dfrac 1{20}$$

$$\dfrac 4{10}$$

$$\dfrac 1{21}$$

$$\dfrac 3{20}$$

Solution

Question 5

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

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

$$\dfrac{7}{15}$$

$$\dfrac{1}{15}$$

Solution

Question 6

$$0.009$$

$$0.047$$

$$0.026$$

None of these

Solution

Question 7

$$\dfrac{19}{144}$$

$$\dfrac{25}{216}$$

$$\dfrac{125}{216}$$

$$\dfrac{175}{216}$$

Solution

Question 8

$$\dfrac{1}{8}$$

$$\dfrac{3}{8}$$

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

$$\dfrac{1}{4}$$

Solution

Question 9

$$\displaystyle \frac{1}{3}$$

$$\displaystyle \frac{1}{2}$$

$$\displaystyle \frac{2}{3}$$

$$\displaystyle \frac{3}{4}$$

Solution

Question 10

$$\left( 200,\ \cfrac { 1 }{ 9 } ,\ \cfrac { 8 }{ 9 } \right) $$

$$\left( 200,\ \cfrac { 2 }{ 9 } ,\ \cfrac { 7 }{ 9 } \right) $$

$$\left( 200,\ \cfrac { 4 }{ 9 } ,\ \cfrac { 5 }{ 9 } \right) $$

$$\left( 200,\ \cfrac { 1 }{ 4 } ,\ \cfrac { 3 }{ 4 } \right) $$

Solution

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