Probability Test

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Probability Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

If $$4P(A) = 6P(B) = 10 P(A \cap B) = 1$$ then $$P\left(\dfrac{B}{A}\right)=$$ _________

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

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

C
$$\dfrac{7}{10}$$

D
$$\dfrac{19}{60}$$

Solution

Let $$4P(A)=6P(B)=10P(A\cap B)=x$$
$$\implies P(A)=\dfrac{x}{4}$$
$$P(B)=\dfrac{x}{6}$$
and
$$P(A\cap B)=\dfrac{x}{10}$$
$$P\left(\dfrac{B}{A}\right)=\dfrac{P(A\cap B)}{P(A)}$$
$$=\dfrac{x/10}{x/4}$$
$$=\dfrac{2}{5}$$
Question 2
1 / -0

Two boys are asked to select each one number from $$1$$ to $$100$$, the probability that they select the same number is

A
$$\dfrac{99}{100}$$

B
$$\dfrac{1}{100}$$

C
$$\dfrac{^{100}C_2}{100^2}$$

D
$$\dfrac{100\times 99}{^{100}C_2}$$

Solution

Let first man select x and it's probability is definite can be any number no chance for particular number
So second man chooses same value x is 1/100
Question 3
1 / -0

Bag $$A$$ contains $$2$$ white and $$3$$ red balls and bag $$B$$ contains $$4$$ white and $$5$$ red balls. One ball is drawn at random from one of the bag is found to be red. Find the probability that it was drawn from bag $$B$$.

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

B
$$\dfrac{25}{52}$$

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

D
$$\dfrac{3}{14}$$

Solution

Let $$X$$ be the probability of choosing bag $$A$$,$$Y$$ be the probability of choosing bag $$B$$

Let $$E$$ be the probability of ball drawn is red

Then $$P\left( X \right) = \dfrac{1}{2}$$

$$P\left( Y \right) = \dfrac{1}{2}$$

$$P\left( {E/X} \right) = \dfrac{3}{5}$$

$$P\left( {E/Y} \right) = \dfrac{5}{9}$$

Apply the Bayes theorem:

$$P\left( {Y/E} \right) = \dfrac{{P\left( Y \right) \times P\left( {E/Y} \right)}}{{P\left( X \right) \times P\left( {E/X} \right) + P\left( Y \right) \times P\left( {E/Y} \right)}}$$

$$ = \dfrac{{\dfrac{1}{2} \times \dfrac{5}{9}}}{{\dfrac{1}{2} \times \dfrac{3}{5} + \dfrac{1}{2} \times \dfrac{5}{9}}}$$

$$ = \dfrac{{50}}{{104}}$$

$$ = \dfrac{{25}}{{52}}$$

Hence, the probability that the red ball is drawn from bag $$B$$ is $$ = \dfrac{{25}}{{52}}$$
Question 4
1 / -0

An urn contains $$10$$ balls coloured either black or red When selecting two balls from the urn at random, the probability that a ball of each color is selected is $$8/15$$. Assuming that the urn contains more black balls then red balls, the probability that at least one black ball is selected, when selecting two balls, is

A
$$\dfrac {18}{45}$$

B
$$\dfrac {30}{45}$$

C
$$\dfrac {39}{45}$$

D
$$\dfrac {41}{45}$$

Solution

Number of black balls $$=x$$

Number of red balls $$=y$$

$$i)=x+y=10$$ (Total)

$$ii)\ P$$ (selecting exactly $$1$$ black & one red)

$$=^xC_1\times ^yC_1 /^{10}C_2=8/15$$

by equation : $$x^2-10x+24=0\ ;\ x=4$$ or $$6$$

since $$ x > y$$ it is $$6$$

$$iii)\ P$$ (selecting at least one black)

$$=^xC_1\times ^yC_1 +^xC_2 /^{10}C_2$$

Above equation reduced to $$\Rightarrow \ 2xy+x(x-1)/90$$

putting $$x=6$$, results is $$39/45$$
Question 5
1 / -0

A random variable $$X$$ has the probability distribution,
$$X = x$$ $$-2$$ $$-1$$ $$0$$ $$1$$ $$2$$ $$3$$
$$P (x)$$ $$\dfrac{1}{10}$$ $$K$$ $$\dfrac{1}{5}$$ $$2K$$ $$\dfrac{3}{10}$$ $$K$$

A
$$\dfrac{2}{10}$$

B
$$\dfrac{1}{10}$$

C
$$\dfrac{3}{10}$$

D
$$\dfrac{7}{10}$$

Solution

The sum of probability of all the possible outcomes should be $$1$$.

That is $$\sum P(x_i)=1$$

$$\implies \dfrac{1}{10}+K+\dfrac{1}{5}+2K+\dfrac{3}{10}+K=1$$

$$\implies 4K+\dfrac{6}{10}=1$$

$$\implies 4K=1-\dfrac{6}{10}=\dfrac{4}{10}$$

$$\implies K=\dfrac{1}{10}$$
Question 6
1 / -0

Three numbers are chosen at random from the set of first $$20$$ numbers, then the probability that their product is a multiple of $$3$$ is

A
$$\frac { 194 }{ 285 } $$

B
$$\frac { 1 }{ 57 } $$

C
$$\frac { 13 }{ 19 } $$

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

Solution
Question 7
1 / -0

If $$P(A)=0.40,P(B)=0.35$$ and $$P\left( A\cup B \right) =0.55$$, then $$P(A/B)=$$ ____

A
$$\cfrac { 1 }{ 5 } $$

B
$$\cfrac { 8 }{ 11 } $$

C
$$\cfrac { 4 }{ 7 } $$

D
$$\cfrac { 3 }{ 4 } $$

Solution

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$P(A\cup B)=0.55$$
$$P(A)=0.40$$
$$P(B)=0.35$$
$$0.55=0.40+0.35-P(A\cap B)$$
$$P(A\cap B)=0.4+0.35-0.55=0.2$$
Now
$$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$$
$$P(A|B)=\dfrac{0.2}{0.35}$$
$$P(A|B)=\dfrac{4}{7}$$
Question 8
1 / -0

A pair of numbers is picked up randomly (without replacement ) from the set $$\{1,2,3,5,7,11,12,13,17,19\}$$. The probability that the number $$11$$ was picked given that the sum of the number was even is nearly:

A
$$0.1$$

B
$$0.125$$

C
$$0.24$$

D
$$0.18$$

Solution

Let, $$A$$ be the event of picking $$11$$ from the set, say $$S$$
Let, $$B$$ be the event where the sum is even. This can happen only when the two numbers picked are both odd or both even.

$$\therefore n(B)=1+_{2}^{8}\textrm{C}1+28=29$$

$$n(A \cap B)=\ _{1}^{7}\textrm{C}$$

Hence, the given probability,

$$P\left(\dfrac{A}{B}\right)=\dfrac{P(A \cap B)}{P(B)}$$

$$P\left(\dfrac{A}{B}\right)=\dfrac{\frac{n(A \cap B)}{n(S)}}{\frac{n(B)}{n(S)}}=\dfrac{n(A \cap B)}{n(B)}=\dfrac{7}{29}\approx\ 0.24$$

The probability that the number $$11$$ was picked such that the number is even is approximately $$0.24$$
Option C
Question 9
1 / -0

One bag contains $$3$$ white & $$2$$ blacks and another contains $$2$$ white & $$3$$ black balls. A ball is drawn from the second bag & placed in the first, then a ball a ball is drawn from the first bag & placed in the second. when the pair of the operations is repeated, the probability that the first bag will contain $$5$$ white balls is:

A
$$1/25$$

B
$$1/125$$

C
$$1/225$$

D
$$2/15$$

Solution
Question 10
1 / -0

Two different dice are tossed together. Find the probability of the sum of no's appearing on two dice is $$5$$.

A
$$\dfrac{1}{36}$$

B
$$\dfrac{1}{18}$$

C
$$\dfrac{1}{9}$$

D
$$\dfrac{1}{6}$$

Solution

$$\vec{b}=(\sqrt{2})a$$
$$(6+2a)||\vec{b}$$
$$n(s)=6\times 6$$
$$n(s)=36$$
$$n(E)=$$ sum $$=5$$
$$=(1, 4), (2, 3), (3, 2), (4, 1)=4$$
$$p(E)=\dfrac{n(E)}{n(s)}=\dfrac{4}{36}=\dfrac{1}{9}$$.

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Re-Attempt Weekly Quiz Competition

$$X = x$$	$$-2$$	$$-1$$	$$0$$	$$1$$	$$2$$	$$3$$
$$P (x)$$	$$\dfrac{1}{10}$$	$$K$$	$$\dfrac{1}{5}$$	$$2K$$	$$\dfrac{3}{10}$$	$$K$$

Probability Test - 36

Result

Probability Test - 36

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SHARING IS CARING

Question 1

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

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

$$\dfrac{7}{10}$$

$$\dfrac{19}{60}$$

Solution

Question 2

$$\dfrac{99}{100}$$

$$\dfrac{1}{100}$$

$$\dfrac{^{100}C_2}{100^2}$$

$$\dfrac{100\times 99}{^{100}C_2}$$

Solution

Question 3

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

$$\dfrac{25}{52}$$

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

$$\dfrac{3}{14}$$

Solution

Question 4

$$\dfrac {18}{45}$$

$$\dfrac {30}{45}$$

$$\dfrac {39}{45}$$

$$\dfrac {41}{45}$$

Solution

Question 5

$$\dfrac{2}{10}$$

$$\dfrac{1}{10}$$

$$\dfrac{3}{10}$$

$$\dfrac{7}{10}$$

Solution

Question 6

$$\frac { 194 }{ 285 } $$

$$\frac { 1 }{ 57 } $$

$$\frac { 13 }{ 19 } $$

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

Solution

Question 7

$$\cfrac { 1 }{ 5 } $$

$$\cfrac { 8 }{ 11 } $$

$$\cfrac { 4 }{ 7 } $$

$$\cfrac { 3 }{ 4 } $$

Solution

Question 8

$$0.1$$

$$0.125$$

$$0.24$$

$$0.18$$

Solution

Question 9

$$1/25$$

$$1/125$$

$$1/225$$

$$2/15$$

Solution

Question 10

$$\dfrac{1}{36}$$

$$\dfrac{1}{18}$$

$$\dfrac{1}{9}$$

$$\dfrac{1}{6}$$

Solution

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