Probability Test

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Probability Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

If $$\dfrac{1+4p}{p},\dfrac{1-p}{4}, \dfrac{1-2p}{2}$$ are propabilities of three mutually exclusive events, then-

A
$$\dfrac{-1}{5}\le p\le \dfrac{1}{2}$$

B
$$\dfrac{1}{3}\le p\le \dfrac{2}{3}$$

C
$$\dfrac{1}{6}\le p\le \dfrac{1}{2}$$

D
None of these

Solution
Question 2
1 / -0

If $$P(A)=\dfrac{1}{2}$$, $$P(B)=\dfrac{3}{8}$$ and $$P(A\cap B)=\dfrac{1}{5}$$ then P(B|A) is equal tp:

A
$$\frac{2}{5}$$

B
$$\frac{8}{15}$$

C
$$\frac{1}{3}$$

D
$$\frac{5}{8}$$

Solution

Given:-
$$P{\left( A \right)} = \cfrac{1}{2}$$
$$P{\left( B \right)} = \cfrac{3}{8}$$
$$P{\left( A \cap B \right)} = \cfrac{1}{5}$$
As we know that,
$$P{\left( B | A \right)} = \cfrac{P{\left( B \cap A \right)}}{P{\left( A \right)}}$$
$$\therefore P{\left( B | A \right)} = \cfrac{\left( \cfrac{1}{5} \right)}{\left( \cfrac{1}{2} \right)}$$
$$\Rightarrow P{\left( B | A \right)} = \cfrac{2}{5}$$
Hence the correct answer is $$\cfrac{2}{5}$$.
Question 3
1 / -0

For the probability distribution given by $$\left.\begin{matrix} X=x_i & 0 \\ P. & \dfrac{25}{36}\end{matrix}\right|$$ $$\begin{matrix} 1 \\ 5 \\ 18\end{matrix}$$ $$\begin{vmatrix} 2 \\ 1 \\ 36\end{vmatrix}$$ the standard deviation $$(\sigma)$$ is?

A
$$\sqrt{\dfrac{1}{3}}$$

B
$$\dfrac{1}{3}\sqrt{\dfrac{5}{2}}$$

C
$$\sqrt{\dfrac{5}{36}}$$

D
None of the above

Solution

$$M^2+\sigma^2=\sum x^2_iP(x=x_i)$$

$$M=\sum x_iP(x=x_i)=\dfrac{5}{18}+\dfrac{2}{36}=\dfrac{6}{18}=\dfrac{1}{3}$$

$$\therefore\dfrac{1}{9}+\sigma^2=\dfrac{5}{18}+\dfrac{4}{36}=\dfrac{7}{18}$$

$$\therefore \sigma^2=\dfrac{7}{18}-\dfrac{2}{18}=\dfrac{5}{18}$$

$$\therefore S.D=\dfrac{1}{3}\sqrt{\dfrac{5}{2}}$$.
Question 4
1 / -0

The probability of happening of an event A is $$0.5$$ and that of B is $$0.3$$. If A and B are mutually exclusive events, then the probability of neither A nor B is?

A
$$0.4$$

B
$$0.5$$

C
$$0.2$$

D
$$0.9$$

Solution

$$P(A)=0.5$$; $$P(B)=0.3$$

$$\therefore P(\bar{A}\cap\bar{B})=1-P(A\cup B)=1-[P(A)+P(B)]$$ $$[\because\ P(A\cap B)=\phi $$ as they are mutually exclusive events $$]$$

$$=1-[0.5+0.3]$$

$$=0.2$$
Question 5
1 / -0

Given that $$A \subset B$$, then identify the correct statement

A
$$P(A/B) = P(A)$$

B
$$P (A/B) \le P(A)$$

C
$$P(A/B) \ge P(A)$$

D
$$P(A/B) = P(A) - P(B)$$

Solution

$$A \subset B$$
$$P (A/B) = \dfrac{P(A \cap B)}{P (B)} = \dfrac{P (A)}{P(B)} \neq P(A)$$ (always)
$$P(A/B) = \dfrac{P(A)}{P(B)} \ge P(A)$$
Question 6
1 / -0

The number of committees formed by taking $$5men$$ and $$5women$$ from $$6women$$ and $$7men$$ are

A
252

B
125

C
126

D
64

Solution
Question 7
1 / -0

Probability of hitting a target independently of $$4$$ persons are $$\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}, \dfrac{1}{8}$$. Then the probability that target is hit, is?

A
$$\dfrac{1}{192}$$

B
$$\dfrac{5}{192}$$

C
$$\dfrac{25}{32}$$

D
$$\dfrac{7}{32}$$

Solution

$$P(H)=1-P$$(Not Hitting)
$$=1-\dfrac{1}{2}\cdot \dfrac{2}{3}\cdot \dfrac{3}{4}\cdot \dfrac{7}{8}=\dfrac{25}{32}$$.
Question 8
1 / -0

A coin is rolled n times. If the probability of getting head at least once is greater than $$90\%$$ then the minimum value of n is?

A
$$4$$

B
$$3$$

C
$$5$$

D
$$6$$

Solution

$$1-\dfrac{1}{2^n} > \dfrac{9}{10}\Rightarrow \dfrac{1}{10} > \dfrac{1}{2^n}\Rightarrow 2^n > 10$$
$$\therefore$$ minimum value of n is $$4$$.
Question 9
1 / -0

Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is:

A
$$\dfrac{1}{11}$$

B
$$\dfrac{1}{17}$$

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

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

Solution

$$P(B) = P(G) = \dfrac{1}{2}$$
Required Probability $$= \dfrac{\text{all 4 girls}}{\text{(all 4 girls) + (exactly 3 giels + 1 boy) + (exactly 2 girls + 2 boys)}}$$

$$= \dfrac{\left(\dfrac{1}{2}\right)^4}{\left(\dfrac{1}{2}\right)^4 + {^4C_3} \left(\dfrac{1}{2}\right)^4 + {^4C_2} \left(\dfrac{1}{2}\right)^4} = \dfrac{1}{11}$$
Question 10
1 / -0

A problem in mathematics is given to $$4$$ students whose chances of solving individually are $$\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}$$ and $$\dfrac{1}{5}$$. The probability that the problem will be solved at least by one student is?

A
$$\dfrac{2}{3}$$

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

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

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

Solution

Given $$P(A)=\dfrac 12,P(A^C)=\dfrac 12$$

$$P(B)=\dfrac 13,P(B^C)=\dfrac 23$$

$$P(C)=\dfrac 14,P(C^C)=\dfrac 34$$

$$P(D)=\dfrac 15,P(D^C)=\dfrac 45$$

$$P(A\cup B\cup C\cup D)=1-P(A^C\cap B^C\cap C^C\cap D^C)$$

$$=1-P(A^C)\cdot P(B^C)\cdot P(C^C)\cdot P(D^C)$$

$$=1-\dfrac{1}{2}\times \dfrac{2}{3}\times \dfrac{3}{4}\times \dfrac{4}{5}$$

$$=1-\dfrac{1}{5}=\dfrac{4}{5}$$.

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Probability Test - 42

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Probability Test - 42

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Question 1

$$\dfrac{-1}{5}\le p\le \dfrac{1}{2}$$

$$\dfrac{1}{3}\le p\le \dfrac{2}{3}$$

$$\dfrac{1}{6}\le p\le \dfrac{1}{2}$$

None of these

Solution

Question 2

$$\frac{2}{5}$$

$$\frac{8}{15}$$

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

$$\frac{5}{8}$$

Solution

Question 3

$$\sqrt{\dfrac{1}{3}}$$

$$\dfrac{1}{3}\sqrt{\dfrac{5}{2}}$$

$$\sqrt{\dfrac{5}{36}}$$

None of the above

Solution

Question 4

$$0.4$$

$$0.5$$

$$0.2$$

$$0.9$$

Solution

Question 5

$$P(A/B) = P(A)$$

$$P (A/B) \le P(A)$$

$$P(A/B) \ge P(A)$$

$$P(A/B) = P(A) - P(B)$$

Solution

Question 6

252

125

126

64

Solution

Question 7

$$\dfrac{1}{192}$$

$$\dfrac{5}{192}$$

$$\dfrac{25}{32}$$

$$\dfrac{7}{32}$$

Solution

Question 8

$$4$$

$$3$$

$$5$$

$$6$$

Solution

Question 9

$$\dfrac{1}{11}$$

$$\dfrac{1}{17}$$

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

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

Solution

Question 10

$$\dfrac{2}{3}$$

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

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

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

Solution

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