Probability Test

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Probability Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

Directions For Questions

A bag contains $$18$$ balls out of which $$x$$ balls are red.
If one ball is drawn at random from the bag, what is the probability that it is not red?

...view full instructions

If the value of x is $$6$$

A
$$\dfrac 23$$

B
$$\dfrac 13$$

C
$$\dfrac 56$$

D
$$\dfrac 16$$

Solution

Total no. of balls $$18$$
Total no.of balls $$6$$
The probability of picking a ball which is not red is $$1-\dfrac6{18}=\dfrac 23$$
Question 2
1 / -0

Directions For Questions

$$8$$ players compete in a tournament, every one plays everyone else just once. The winner of a game gets $$1$$, the loser $$0$$ or each gets $$\dfrac{1}{2}$$ if the game is drawn. The final result is that every one gets a different score and the player playing placing second gets the same as the total of four bottom players.

...view full instructions

The score of the player placing II was

A
$$5\dfrac{1}{2}$$

B
$$6\dfrac{1}{2}$$

C
$$6$$

D
$$can't\ say$$

Solution

The total no.of matches held is given by $$\dfrac{8\times 7}{2}=28$$
So the winner is the one who has the score in between $$[28,14]$$
The second is the one who has less score than the first one.
Since it has many possibilities it cannot be determined.
Question 3
1 / -0

A fair die is thrown 3 times . The chance that sum of three numbers appearing on the die is less than 11 , is equal to -

A
$$\dfrac{1}{2}$$

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

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

D
$$\dfrac{5}{8}$$

Solution

From given, we have,

Sum 3:$$ (1, 1, 1)$$ ==> Contributing only $$1$$ distinct triplet.

Sum 4: $$(1, 1, 2)$$ ==> Contributing $$3$$ distinct triplets

Sum 5: $$(1, 2, 2) $$and $$(1, 1, 3)$$ ==> Contributing $$6$$ distinct triplets

Sum 6: $$(1, 1, 4), (1, 2, 3)$$ and $$(2, 2, 2)$$ => Together contributing $$10 $$distinct triplets

Sum 7: $$(1, 1, 5), (1, 2, 4), (1, 3, 3) $$and $$(2, 2, 3)$$ => Together contributing $$15$$ distinct triplets.

Sum 8: $$(1, 1, 6), (1, 2, 5), (1, 3, 4), (2, 3, 3)$$ and$$ (2, 4, 2) $$==> Together contributing $$21$$ distinct triplets.

Sum 9: $$(1, 2, 6), (1, 3, 5), (1, 4, 4), (2, 3, 4), (2, 5, 2)$$ and $$(3, 3, 3)$$ ==> Together contributing $$25$$ distinct triplets.

Sum 10: $$(1, 3, 6), (1, 4, 5), (2, 2, 6), (2, 3, 5), (2, 4, 4)$$ and $$(3, 3, 4) $$==> Together contributing $$27$$ distinct triplets.

Therefore number of favorable cases $$= 1+ 3 + 6 + 10 + 15 + 21 + 25 + 27 = 108.$$

Therefore, probability $$= \dfrac{108}{216} =\dfrac{1}{2}$$
Question 4
1 / -0

A,B and C are three mutually exclusive and exhaustive events and $$P(B)=\dfrac{3}{2}P(A), P(C)=\dfrac{1}{3}P(B)$$ then the value of $$P(A)$$ is

A
1/3

B
1/2

C
1/6

D
none of these

Solution

Given, $$A, B$$ and $$C$$ are three mutually exclusive and exhaustive events and $$P(B)=\dfrac{3}{2}P(A), P(C)=\dfrac{1}{3}P(B)$$.

Since $$A, B, C$$ are mutually exclusive and exhaustive events then we've,

$$P(A)+P(B)+P(C)=1$$

or, $$P(A)+\dfrac{3}{2}P(A)+\dfrac{1}{2}P(A)=1$$

or, $$3P(A)=1$$

or, $$P(A)=\dfrac{1}{3}$$.
Question 5
1 / -0

From a well shuffled deck of $$52$$ cards, $$4$$ cards are drawn at random. What is the probability that all the drawn cards are of the same colour.

A
$$\dfrac{184}{833}$$

B
$$\dfrac{92}{833}$$

C
$$\dfrac{92}{2466}$$

D
$$\dfrac{46}{833}$$

Solution
Question 6
1 / -0

The probability that a number selected at random from the numbers $$1,2,3.......15$$ is a multiple of $$4$$ is

A
$$\dfrac{4}{15}$$

B
$$\dfrac{2}{15}$$

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

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

Solution

From Number$$ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15$$

From 1 to 15, Multiples of 4 are 4, 8, 12 only

So Probability$$ = \dfrac {Count \ of \ No. \ which \ are \ multiple \ of \ 4}{Total \ Number \ given}$$

$$ Prob. = \dfrac{3}{15} = \dfrac{1}{5}$$

Option D is correct
Question 7
1 / -0

If $$A$$ and $$B$$ are events such that $$P(A)=0.3,P(B)=0.2$$ and $$P(\bar { A } \cup B)=$$?

A
$$0.2$$

B
$$0.1$$

C
$$0.4$$

D
$$0.5$$

Solution

$$B=\left( \bar { A } \cup A \right) \cap B=\left( \bar { A } \cap B \right) \cup \left( A\cap B \right) $$

$$\Rightarrow P(B)=P\left( \bar { A } \cap B \right) +P\left( A\cap B \right) $$

$$\Rightarrow P\left( \bar { A } \cap B \right) =P(B)-P\left( A\cap B \right) =(0.2-0.1)$$

$$\Rightarrow P\left( \bar { A } \cap B \right) =0.1\quad $$
Question 8
1 / -0

If $$A$$ and $$B$$ are events at the same experiments with $$P(A)=0.2, P(B)=0.5$$ then maximum value of $$P(A'\cap B)$$ is

A
$$1/4$$

B
$$1/2$$

C
$$1/8$$

D
$$1/16$$

Solution

A and B are experiments with $$P(A)=0.2$$ and $$P(B)=0.5$$.
Thus, $$P(A')=1-P(A)=1-0.2=0.8$$
Now the max value of $$P(A'\bigcap B)$$ is the lower of the probabilities $$P(A')$$ and $$P(B)$$.
Since, $$P(B) < P(A')$$ hence max $$P(A' \bigcap B)=P(B)=0.5-\frac{1}{2}$$

Thus, the correct answer is option B.
Question 9
1 / -0

If the probability for $$A$$ to fail in an examination is $$0.2$$ and that for $$B$$ is $$0.30$$, then the probability that either $$A$$ or $$B$$ fails, is

A
$$0.38$$

B
$$0.44$$

C
$$0.50$$

D
$$0.94$$

Solution
Question 10
1 / -0

A students appears for tests I,II and III.The student is successful if he passes either in tests I and II or tests I and III.The probabilities of the student passing in test I,II and III are respectively, $$p,q,$$ and $$1/2$$.If the probability that the student is successful is $$1/2,$$ then $$p(1+q)=$$

A
1/2

B
1

C
3/2

D
3/4

Solution

If $$A,B,C$$ represent events that the students is successful in tests $$I,II,III$$ respectively.
$$P(A)=p,P(B)=q,P(C)=\dfrac12,P(S)=\dfrac12$$
Then the probability that the student is successful is
$$P(S)=P[(A \cap B \cap C') \cup (A \cap B' \cap C)\cup (A \cap B\cap C)]$$
$$=P(A)P(B)P(C')+P(A)P(B')P(C)+P(A)P(B)P(C)$$
................ [$$\therefore,A,B,C$$ are independent events]
$$=pq(1-\dfrac{1}{2})+p(1-q)\dfrac{1}{2}+pq\dfrac{1}{2}$$
$$=pq+\dfrac{1}{2}p-\dfrac{1}{2}pq$$
$$=\dfrac{1}{2}(pq+p)\\$$
$$\therefore \dfrac{1}{2}p(1+q)=\dfrac{1}{2}$$
$$\Rightarrow p(1+q)=1$$

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Re-Attempt Weekly Quiz Competition

Probability Test - 51

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Probability Test - 51

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SHARING IS CARING

Question 1

$$\dfrac 23$$

$$\dfrac 13$$

$$\dfrac 56$$

$$\dfrac 16$$

Solution

Question 2

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

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

$$6$$

$$can't\ say$$

Solution

Question 3

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

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

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

$$\dfrac{5}{8}$$

Solution

Question 4

1/3

1/2

1/6

none of these

Solution

Question 5

$$\dfrac{184}{833}$$

$$\dfrac{92}{833}$$

$$\dfrac{92}{2466}$$

$$\dfrac{46}{833}$$

Solution

Question 6

$$\dfrac{4}{15}$$

$$\dfrac{2}{15}$$

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

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

Solution

Question 7

$$0.2$$

$$0.1$$

$$0.4$$

$$0.5$$

Solution

Question 8

$$1/4$$

$$1/2$$

$$1/8$$

$$1/16$$

Solution

Question 9

$$0.38$$

$$0.44$$

$$0.50$$

$$0.94$$

Solution

Question 10

1/2

1

3/2

3/4

Solution

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