Free Objective Test 02 Practice Test - 11th and 12th

Question 1

A set of integers is given as (3,6,8,14,17). What is the probability that a triangle can be constructed.?

\frac{3}{5}

\frac{3}{10}

\frac{1}{10}

\frac{5}{10}

SOLUTION

Solution : B

Any 3 points can be selected in $^{5} C_{3}$ i.e 10 ways.

For forming a triangle sum of two sides should be greater than the $3^{r d}$ side.

Hence following set of integers can be selected

(3,6,8)

(6,8,17)

(8,14,17)

Hence P(Triangle is formed)= $\frac{3}{10}$

Question 2

Letters of the word "EDUCATION" is arranged. What is the probability that vowels and consonants are in alphabetical order?

\frac{1}{(5! \times 4!)}

\frac{1}{9!}

\frac{9}{9!}

D. None of these.

SOLUTION

Solution : A

There are 5 vowels and 4 consonants.

Total ways to arrange: 9!

Vowels can be arranged in 5! ways out of which in $\frac{1}{5!}$ they are in alphabetical order.

Similarly for consonants.

Total Favorable ways: $\frac{9!}{(5! \times 4!)}$

Total ways: 9!

P(E)= $\frac{1}{(5! \times 4!)}$

Question 3

There are 4 torn books in a lot consisting of 10 books. If 5 books are seleceted at random what is the probability of presence of 2 torn books among the selected?

\frac{1}{3}

\frac{4}{21}

\frac{10}{21}

\frac{2}{7}

SOLUTION

Solution : C

Total Sample space: $^{10} C_{5} = 252$

Favorable cases: $^{4} C_{2} \times^{6} C_{3} = 120$

Hence P(E)= $\frac{120}{252} = \frac{10}{21}$

Question 4

Of the 10 prizes 5 prizes are of category Platinum 3 of gold and 2 of silver and they are placed in an enclosure for an olympiad contest. The prizes are awarded by allowing winners to select randomly from the prizes remaining. When the 8th participant goes to collect the prize what the probability that last 3 prizes are 1 of platinum 1 of gold and 1 of silver?

\frac{1}{5}

\frac{1}{4}

\frac{1}{3}

\frac{1}{10}

SOLUTION

Solution : B

Sample space: $^{10} C_{7}$ . because 7 prizes have already been selected= 120.

Favorable: $^{5} C_{4} \times^{3} C_{2} \times^{2} C_{1} = 30$ .

P(E)= $\frac{30}{120} = \frac{1}{4}$

Question 5

A fortune teller has numbers from 1-7 in his box. He draws 3 numbers from it. What is the probability that the number picked is of alternate order as odd-even-odd or even-odd-even?

\frac{1}{14}

\frac{2}{7}

\frac{9}{14}

D. None of the above.

SOLUTION

Solution : B

Sample Space: $7 \times 6 \times 5 = 210$

odd-even-odd= $4 \times 3 \times 3 = 36$

even-odd-even= $3 \times 4 \times 2 = 24$

Total ways: 60

P(E)= $\frac{60}{210} = \frac{2}{7}$

Question 6

Two cubes have their face painted as either red or blue color. 1st cube has 5 faces red and 1 face blue. The Probability that both the faces show same color on rolling is $\frac{1}{2}$ . How many red faces are present on the $2^{n d}$ cube.

A. 6

B. 4

C. 3

D. 2

SOLUTION

Solution : C

Suppose x faces of cube 2 is red then 6-x will be blue.

P(RR or BB)= $\frac{5}{6} \times \frac{x}{6} + \frac{1}{6} \times \frac{(6 - x)}{6} = \frac{1}{2}$

$\frac{5 x}{36} + \frac{(6 - x)}{36} = \frac{1}{2}$

4x+6=18

x=3

Hence number of red faces is 3.

Question 7

The probability that a 'P and C' question will be asked in IIT JEE is $\frac{2}{5}$ and probability question is uploaded is $\frac{4}{7}$ . If the probability of getting at least 1 is $\frac{2}{3}$ what is the probability that questions from both the topics are asked.

\frac{17}{105}

\frac{16}{105}

\frac{1}{35}

\frac{6}{35}

SOLUTION

Solution : A

P(A U B)= P(A)+ P(B)- P(A intersection B)

$\frac{2}{3} = (\frac{2}{5}) + (\frac{4}{7}) - x$

$x = \frac{17}{105}$

Question 8

A maze with following pattern is given. An insect has to reach point R.It can only land only on R if it lands on the adjacent cubes. What is the probability of it reaching R.

\frac{1}{4}

\frac{2}{3}

\frac{2}{7}

\frac{1}{6}

SOLUTION

Solution : A

P(It lands on R)= P( S to R)+ P(Q to R)

From Q it can move in 2 directions.

P(Landing on R)= $\frac{1}{6} \times 1 + \frac{1}{6} \times \frac{1}{2} = \frac{1}{4}$

Question 9

CSK lost the toss 13 out of 14 times. What is the probability of this event?

\frac{3}{2^{14}}

\frac{5}{2^{14}}

\frac{7}{2^{13}}

D. None of the above

SOLUTION

Solution : C

13 losses from 14 can be selected in $^{14} C_{13}$ ways i.e.14

At any of the tosses only 2 events can happen win or loss hence total sample space: $2^{14}$

P(E)= $\frac{14}{2^{14}}$

Question 10

A bag contains 'x' red balls '2x' white balls and '3x' black balls. 3 balls are drawn at random. The probability that all the balls drawn are of different colors is 0.3 .How many white balls are present in the bag?

A. 2

B. 4

C. 5

D. 7

SOLUTION

Solution : A

P(All different)= $(^{x} C_{1}) \times (^{2 x} C_{1}) \times (^{3 x} C_{1}) / (^{6 x} C_{3})$

P(E)= $(6 x^{3}) \times \frac{6}{6 x} \times [(6 x - 1) \times (6 x - 2)] = \frac{3}{10}$

or, $\frac{3 x^{2}}{[(6 x - 1) \times (6 x - 2)]} = \frac{3}{10}$

$10 x^{2} = 18 x^{2} - 9 x + 1$

or, (x-1)(8x-1)=0

x can not be a fraction hence x=1

Total number of white balls= 2x=2

Question 11

Urn A contains 6 red and 4 black balls and urn B contains 4 red and 6 black balls. One ball is drawn at random from urn A and placed in urn B. Then one ball is drawn at random from urn B and placed in urn A. If one ball is now drawn at random from urn A, the probability that it is found to be red, is

\frac{32}{55}

\frac{21}{55}

\frac{19}{55}

D. None of these

SOLUTION

Solution : A

Let the events are
$R_{1}$ = A red ball is drawn from urn A and placed in B
$B_{1}$ = A black ball is drawn from urn A and placed in B
$R_{2}$ = A red ball is drawn from urn B and placed in A
$B_{2}$ = A black ball is drawn from urn B and placed in A
R = A red ball is drawn in the second attempt from A

Then the required probability

= $P (R_{1} R_{2} R) + (R_{1} B_{2} R) + P (B_{1} R_{2} R) + P (B_{1} B_{2} R)$
= $P (R_{1}) P (R_{2}) P (R) + P (R_{1}) P (B_{2}) P (R) + P (B_{1}) P (R_{2}) P (R) + P (B_{1}) P (B_{2}) P (R)$
= $\frac{6}{10} \times \frac{5}{11} \times \frac{6}{10} + \frac{6}{10} \times \frac{6}{11} \times \frac{5}{10} + \frac{4}{10} \times \frac{4}{11} \times \frac{7}{10} + \frac{4}{10} \times \frac{7}{11} \times \frac{6}{10}$
= $\frac{32}{55}$

Question 12

The probability of a bomb hitting a bridge is $\frac{1}{2}$ and two direct hits are needed to destroy it. The least number of bombs required so that the probability of the bridge beeing destroyed is greater then 0.9, is

A. 8

B. 7

C. 6

D. 9

SOLUTION

Solution : A

Let n be the least number of bombs required and x the number of bombs that hit the bridge. Then x follows a binomial distribution with parameter n and $p = \frac{1}{2}$
Now, $P (X \geq 2) > 0.9 \Rightarrow 1 - P (X < 2) > 0.9$
$\Rightarrow P (X = 0) + P (X = 1) < 0.1$
$\Rightarrow^{n} C_{0} {(\frac{1}{2})}^{n} +^{n} C_{1} {(\frac{1}{2})}^{n - 1} (\frac{1}{2}) < 0.1 \Rightarrow 10 (n + 1) < 2^{n}$
This gives $n \geq 8$

Question 13

One of the two events must occur. If the chance of one is $\frac{2}{3}$ of the other, then odds in favour of the other are

2 : 3

1 : 3

3 : 1

3 : 2

SOLUTION

Solution : D

Let p be the probability of the other event, then the probability of the first event is $\frac{2}{3}$ p. Since two events are totally exclusive, we have $p + (\frac{2}{3}) p = 1 \Rightarrow p = \frac{3}{5}$
Hence odds in favour of the other are 3 : 5 – 3, i.e, 3 : 2.

Question 14

The probability of happening 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 happening neither A nor B is

A. 0.6

B. 0.2

C. 0.21

D. None of these

SOLUTION

Solution : B

$P (¯ ¯¯ ¯ A \cap ¯ ¯¯ ¯ B) = P (¯ ¯¯ ¯ A \cup ¯ ¯¯ ¯ B) - 1 - P (A \cup B)$
Since A and B are mutually exclusive,
$P (A \cup B) = P (A) + P (B)$
Hence required probability $= 1 - (0.5 + 0.3) = 0.2$

Question 15

If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the form $7^{m} + 7^{n}$ is divisible by 5 equals

\frac{1}{4}

\frac{1}{7}

\frac{1}{8}

\frac{1}{49}

SOLUTION

Solution : A

Since m and n are selected between 1 and 100, hence sample space = $100 \times 100.$
Also $7^{1} = 7, 7^{2} = 49, 7^{3} = 343, 7^{4} = 2401, 7^{5} = 16807$ etc. Hence 1, 3, 7 and 9 will be the last digits in the powers of 7. Hence for favorable cases
$n m \to ↓ 1, 11, 21, 3.... 1, 100 2, 12, 22, 3... 2, 100 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100, 1100, 2100, 3100, 100$
For m = 1; n = 3, 7, 11….. 97
$∴$ Favorable cases = 25
Similarly for every m, favorable n are 25.
$∴$ Total favourable cases = $100 \times 25$
Hence required probability = $\frac{100 \times 25}{100 \times 100} = \frac{1}{4}$

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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