Probability Test

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Probability Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

A die is rolled 5 times. The probability there are at lest two equal numbers among the outcomes obtained is ___________.

A
319/324

B
49/54

C
13/18

D
4/9

Solution
Question 2
1 / -0

A bag contains $$6$$ red, $$4$$ white and $$8$$ blue balls. If three balls are drawn at random, find the probability that one is red, one is white and one is blue.

A
$$\dfrac {2}{17}$$

B
$$\dfrac {3}{17}$$

C
$$\dfrac {5}{17}$$

D
$$\dfrac {4}{17}$$

Solution

$$E\rightarrow$$ Event of getting one is red, one is white and one $$6$$ red $$+4$$ white $$+8$$ blue balls $$=18$$ balls

Total outcomes $$=(\ ^{18}C_{3})$$

$$\underbrace { \bigodot }_{ R } \underbrace { \bigodot }_{ W } \underbrace { \bigodot }_{ B } \rightarrow$$ no. of fobourable element

$$=\ ^{6}C_{1}\times \ ^{4}C_{1}\times \ ^{8}C_{1}$$

$$=6\times 4\times 8$$

$$\therefore P(E)=\dfrac{6\times 4\times 8}{\ ^{18}C_{3}}=\dfrac{6\times 4\times 8\times 3\times 2\times 1}{18\times 17\times 16}$$

$$=\dfrac{4}{17}$$
Question 3
1 / -0

In a random experiment a fair die is rolled until two fours are obtained in succession the probability that the experiment will end in the fifth throw of the die is equal.

A
$$\dfrac {150}{6^{5}$$

B
$$\dfrac {175}{6^{5}}$$

C
$$\dfrac {200}{6^{5}}$$

D
$$\dfrac {225}{6^{5}}$$

Solution

$$\dfrac { 1 } { 6 ^ { 2 } } \left( \dfrac { 5 ^ { 3 } } { 6 ^ { 3 } } + \dfrac { 2 \mathrm { C } _ { 1 } \cdot 5 ^ { 2 } } { 6 ^ { 3 } } \right) = \dfrac { 175 } { 6 ^ { 5 } }$$
Ans-Option B
Question 4
1 / -0

If $$P ( A ) = 0.8,$$ $$P ( B ) = 0.5$$ and $$P \left( \dfrac { B } { A } \right) = 0.4$$ then $$P \left( \dfrac { A } { B } \right) =?$$

A
$$0.32$$

B
$$0.64$$

C
$$0.16$$

D
$$0.25$$

Solution

$$P(A)=0.8 ; P(B)=0.5$$

$$P(\dfrac{B}{A})=0.4$$

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

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

$$\dfrac{P(A\cap B)}{P(A\cap B)}=\dfrac{P(A)\times P(B/A)}{P(B)\times P(A/B)}$$

$$\Rightarrow 1=\dfrac{0.8\times 0.4}{0.5\times P(A/B)}$$

$$\Rightarrow 0.5\times P(A/B)=0.32$$

$$P(A/B)=\dfrac{0.32}{0.5}\Rightarrow 0.64$$
Question 5
1 / -0

Three number are chosen at random without replacement from {1,2,3,...8}. The probability that their minimum is 3, given that their maximum is 6 is

A
$$\frac{3}{28}$$

B
$$\frac{1}{5}$$

C
$$\frac{1}{4}$$

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

Solution
Question 6
1 / -0

The face cards are removed from a full pack. Out of the remaining $$40$$ cards, $$4$$ are drawn at random. What is the probability that they belong to different suits?

A
$$\dfrac {1000}{9139}$$

B
$$\dfrac {1001}{9136}$$

C
$$\dfrac {100}{913}$$

D
$$\dfrac {1002}{9129}$$

Solution
Question 7
1 / -0

Two integers are selected at random from the set $${1,2,...,11}$$. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is:

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

B
$$\dfrac{1}{2}$$

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

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

Solution

Since sum of two numbers is even so either both are odd or both are even. Hence number of elements in reduced samples space $$=^5C_2+^6C_2$$
so required probability $$=\dfrac{^5C_2}{^5C_2+^6C_2}$$ = $$\dfrac{2}{5}$$
Question 8
1 / -0

Three letters are dictated to three persons and an envelop is addressed to each of them, the letters are insearted into the envelop at random so that each envelop contains exactly one letter . Find the probability that atleast one letter is in its proper envelop .

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

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

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

D
None of these

Solution
Question 9
1 / -0

Directions For Questions

In a class of $$10$$ students, probability of exactly $$i$$ students passing an examination is directly proportional to $$i^{2}$$.

...view full instructions

If a students selected at random is found to have passed the examination, then the probability that he was the only students who has passed the examination is

A
$$1/3025$$

B
$$1/605$$

C
$$1/275$$

D
$$1/121$$$

Solution

Let P(i) be the probability that exactly i students are passing an examination.Now given that
$$P(A_i)=\lambda i^2$$ (where $$\lambda$$ is constant)
$$\Rightarrow \sum^{10}_{i=1}P(A_i)=\sum^{10}_{i=1} \lambda \dfrac{10 \times 11\times 21}{6}=\lambda 385=1\Rightarrow \lambda \Rightarrow 1/385$$

Let A represent the event that selected students have passed the examination.
$$\therefore P(A)=\sum^{10}_{i=1}P(A/A_i)P(A_i)$$
$$=\sum^{10}_{i=1}\dfrac{i}{10}\dfrac{i^1}{385}$$
$$=\dfrac{1}{3850}\sum^{10}_{i=1}i^3$$
$$=\dfrac{10^211^2}{4 \times 3850}=\dfrac{11}{14}$$
Now,
$$P(A_1/A)=\dfrac{P(A/A_1)P(A_1)}{P(A)}$$
$$=\dfrac{\dfrac{1}{385}\dfrac{1}{10}}{\dfrac{11}{14}}$$
$$=\dfrac{1}{11\times 55}\dfrac{1}{5}$$
$$=\dfrac{1}{3025}$$
Question 10
1 / -0

Directions For Questions

In a class of $$10$$ students, probability of exactly $$i$$ students passing an examination is directly proportional to $$i^{2}$$.

...view full instructions

If a students is selected random, then the probability that the has passed the examination is

A
$$1/7$$

B
$$11/35$$

C
$$11/14$$

D
$$none\ of\ these$$

Solution

Let P(i) be the probability that exactly i students are passing an examination.Now given that
$$P(A_i)=\lambda i^2$$ (where $$\lambda$$ is constant)
$$\Rightarrow \sum^{10}_{i=1}P(A_i)=\sum^{10}_{i=1} \lambda \dfrac{10 \times 11\times 21}{6}=\lambda 385=1\Rightarrow \lambda \Rightarrow 1/385$$

Let A represent the event that selected students have passed the examination.
$$\therefore P(A)=\sum^{10}_{i=1}P(A/A_i)P(A_i)$$
$$=\sum^{10}_{i=1}\dfrac{i}{10}\dfrac{i^1}{385}$$
$$=\dfrac{1}{3850}\sum^{10}_{i=1}i^3$$
$$=\dfrac{10^211^2}{4 \times 3850}=\dfrac{11}{14}$$