Free Objective Test 01 Practice Test - 11th and 12th

Question 1

The odds in favour of A solving a problem are 3 to 4 and the odds against B solving the same problem are 5 to 7. If they both try the problem, the probability that the problem is solved is:

\frac{41}{84}

\frac{16}{21}

\frac{5}{21}

\frac{1}{4}

SOLUTION

Solution : B

$P (A) = \frac{3}{7}, P (B) = \frac{7}{12} P (¯ A) = \frac{4}{7}, P (¯ B) = \frac{5}{12} P (A \cup B) = 1 - P (¯ ¯¯¯¯¯¯¯¯¯¯¯¯ ¯ A \cup B) = 1 - P (¯ A \cap ¯ B) ∴ P (A \cup B) = 1 - \frac{4}{7} \times \frac{5}{12} = \frac{16}{21}$

Question 2

Two players A and B throw a die alternately for a prize of Rs 11, which is to be won by a player who first throws a six. If A starts the game, their respective expectations are

A. Rs 6; Rs 5

B. Rs 7; Rs 4

C. Rs 5.50; Rs 5.50

D. Rs 5.75; Rs 5.25

SOLUTION

Solution : A

Probability through a six = $\frac{1}{6}$
$P (A) = \frac{1}{6}, P (¯ A) = \frac{5}{6}, P (B) = \frac{1}{6}, P (¯ B) = \frac{5}{6}$
Probability of A winning
$= P (A) + P (¯ A) P (¯ B) P (A) + P (¯ A) P (¯ B) P (¯ A) P (¯ B) P (A) + . . .$
$= \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + . . .$
$= \frac{\frac{1}{6}}{1 - \frac{25}{36}} = \frac{6}{11}$
Probability of B winning $= 1 - \frac{6}{11} = \frac{5}{11}$
$∴$ Expectations of A and B are
$\frac{6}{11} \times 11 =$ Rs 6 and $\frac{5}{11} \times 11$ =Rs 5

Question 3

Sixteen players $P_{1} . P_{2} . \dots \dots \dots P_{16}$ play in a tournament. They are divided into eight pairs at random.From each pair a winner is decided on the basis of a game played between the two players of the pair.Assuming that all the players are of equal strength, the probability that exactly one of the two players $P_{1}$ and $P_{2}$ is among the eight winners is

\frac{4}{15}

\frac{7}{15}

\frac{8}{15}

\frac{17}{30}

SOLUTION

Solution : C

Let $E_{1} (E_{2})$ denote the event that $P_{1}$ and $P_{2}$ are paired (not paired) together and let A denote the event that one of two players $P_{1}$ and $P_{2}$ is among the winners.
Since, P1 can be paired with any of the remaining 15 players.
We have, $P (E_{1}) = \frac{1}{15}$
and $P (E_{2}) = 1 - P (E_{1}) = 1 - \frac{1}{15} = \frac{14}{15}$
In case $E_{1}$ occurs, it is certain that one of $P_{1}$ and $P_{2}$ will be among the winners. In case $E_{2}$ occurs, the probability that exactly of $P_{1}$ and $P_{2}$ is among the winners is
$P {(P_{1} \cap_{2}) \cup (_{1} \cap P_{2})} = P (P_{1} \cap_{2}) + P (_{1} \cap P_{2}) = P (P_{1}) P (_{2}) + P (_{1}) P (P_{2}) = (\frac{1}{2}) (1 - \frac{1}{2}) + (1 - \frac{1}{2}) (\frac{1}{2}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
ie, $P (\frac{A}{E_{1}}) = 1$ and $P (\frac{A}{E_{2}}) = \frac{1}{2}$
By the total probability Rule,
$P (A) = P (E_{1}) . P (\frac{A}{E_{1}}) + P (E_{1}) . P (\frac{A}{E_{2}}) = \frac{1}{15} (1) + \frac{14}{15} (\frac{1}{2}) = \frac{8}{15}$

Question 4

A box contains 24 identical balls of which 12 are white and 12 black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the $4^{t h}$ time on the $7^{t h}$ draw is

\frac{5}{64}

\frac{27}{32}

\frac{5}{32}

\frac{1}{2}

SOLUTION

Solution : C

In any trail, P(getting white ball) = $\frac{1}{2}$
P(getting black ball) = $\frac{1}{2}$
Now, required event will occur if in the first six trails 3 white balls are drawn in any one of the 3 trails from six. The remaining 3 trails must be kept reserved for black balls. This can happen
In $^{6} C_{3} \times^{3} C_{3} = 20$ ways.
$= 20 \times {(\frac{1}{2})}^{3} \times {(\frac{1}{2})}^{3} \times \frac{1}{2} = \frac{5}{32}$

Question 5

Three children are selected at random from a group of 6 boys and 4 girls. It is known that in this group exactly one girl and one boy belong to same parent. The probability that the selected group of children have no blood relations, is equal to

\frac{1}{15}

\frac{13}{15}

\frac{14}{15}

\frac{2}{15}

SOLUTION

Solution : C

P(boyand his sister both are selected) = $\frac{8 C_{1}}{10 C_{3}} = \frac{1}{15}$
$∴$ required probability = $1 - \frac{1}{15} = \frac{14}{15}$

Question 6

A bag contains 3 white, 3 black and 2 red balls. One by one, three balls are drawn without replacing them. Then the probability that the third ball is red , is given by

\frac{5}{24}

\frac{1}{12}

\frac{1}{4}

\frac{1}{2}

SOLUTION

Solution : C

Number of white balls = 3, black balls = 3 red balls = 2
Since drawn balls are not replaced, for third ball to red, we have following patterns:
$E_{1} = W W R, B B R, E_{2} = B W R, W B R; a n d E_{3} = R B R, R W R, B R R, W R R$
$P (E_{1}) = 2 \times \frac{3}{8} \times \frac{2}{7} \times \frac{2}{6}, P (E_{2}) = 2 \times \frac{3}{8} \times \frac{3}{7} \times \frac{2}{6}, P (E_{3}) = \frac{4 \times 2 \times 3 \times 1}{8.7.6}$
Required probability $= P (E_{1}) + P (E_{2}) + P (E_{3}) = \frac{1}{4}$

Question 7

A natural number is chosen at random from the first 100 natural numbers. Then the probability, for the in-equation $x + \frac{100}{x} > 50$ satisfied, is

\frac{1}{20}

\frac{11}{20}

\frac{1}{3}

\frac{1}{4}

SOLUTION

Solution : B

$x + \frac{100}{x} > 50 \Rightarrow x^{2} - 50 x + 100 > 0 \Rightarrow 1 \leq x < 25 - 5 \sqrt{21}, o r 25 + 5 \sqrt{21} < x \leq 100 (1 \leq x \leq 100, x ϵ N)$
$\Rightarrow x = 1, 2, 48, 49, 50, . . ., 100$
Therefore the total number of ways = 100 and favorable ways = 55
Required probability $= \frac{55}{100} = \frac{11}{20} .$

Question 8

The probability that a certain beginner at golf gets good shot if he uses correct club is $\frac{1}{3},$ and the probability of a good shot with an incorrect club is $\frac{1}{4} .$ In his bag there are 5 different clubs only one of which is correct for the good shot. If he chooses a club at random and take a stroke, the probability that he gets a good shot is

\frac{1}{3}

\frac{1}{12}

\frac{4}{15}

\frac{7}{12}

SOLUTION

Solution : C

P( getting correct club ) $= \frac{1}{5}$
Therefore P(hitting good shot by correct club) $= \frac{1}{3} \times \frac{1}{5} = \frac{1}{15}$
P(getting wrong club) $= \frac{4}{5}$
P(hitting good shot with wrong club) $= \frac{4}{5} \times \frac{1}{4} = \frac{1}{5} .$
Therefore P(hitting good shot) $= \frac{1}{15} + \frac{1}{5} = \frac{4}{15} .$

Question 9

A coin whose faces marked 2 and 3 is thrown 5 times, then chance of obtaining a total of 12 is

\frac{5}{16}

\frac{5}{8}

\frac{5}{32}

\frac{5}{24}

SOLUTION

Solution : A

It is given that the coin has faces 2 and 3, tossed five times.

To get sum 12 we need 2,2,2,3,3 in any combination.

So, in the 5 throws, there need to be 3 2's. Automatically other 2 will be 3's since there is no other possibility.

So, it boils down to choosing 3 from 5.

Therefore P(getting sum 12) $= 5 C_{3} {(\frac{1}{2})}^{5} = \frac{5}{16} .$

Question 10

Let `head` means 1 and `tail` means 2 and coefficients of the equation $a x^{2} + b x + c = 0$ are chosen by tossing a fair coin. The probability that the roots of the equation are non-real, is equal to

\frac{5}{8}

\frac{7}{8}

\frac{3}{8}

\frac{1}{8}

SOLUTION

Solution : B

a, b, c may be 1 or 2
$a x^{2} + b x + c = 0$ has non-real roots if $b^{2} - 4 a c < 0$
$\begin{matrix} b & 1 & 2 (a, c) & (1, 1), (1, 2), (2, 1), (2, 2) & (1, 2), (2, 1) (2, 2) \end{matrix} = 7$
Hence required probabilidy $= \frac{7}{2^{3}} = \frac{7}{8} .$

Question 11

Two fair dice are rolled simultaneously. It is found that one of the dice show odd prime number. The probability that the remaining dice also show an odd prime number, is equal to

\frac{1}{5}

\frac{2}{5}

\frac{3}{5}

\frac{4}{5}

SOLUTION

Solution : A

Odd prime on a die are 3 and 5.
Let event A = one of the dice show odd prime number, and event B = remaining dice also shows odd prime number.
$P (\frac{B}{A}) = \frac{n (A ⋂ B)}{n (A)} . A = {(3, 1) (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), . . . (5, 6), (1, 3), (2, 3), . . . (6, 3), (1, 5), . . . (5, 5),} A n d A ⋂ B = {(3, 3) . (5, 5), (3, 5) (5, 3)} ∴ P (\frac{A}{B}) = \frac{4}{20} = \frac{1}{5} .$

Question 12

A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to that getting 9 heads, then the probability of getting 3 heads is

\frac{35}{2^{12}}

\frac{35}{2^{14}}

\frac{7}{2^{12}}

\frac{7}{2^{14}}

SOLUTION

Solution : A

Probability of r successes in n trials is equal to $^{n} C_{r} p^{r} q^{n - r}$ , where p is the probability of success in one trial and q is the probability of failure in one trial.
Let the coin be tossed n times. Here p and q are both 1/2
$P (g e t t i n g 7 h e a d s) = n C_{7} {(\frac{1}{2})}^{n} a n d P (g e t t i n g 9 h e a d s) = n C_{9} {(\frac{1}{2})}^{n} G i v e n, P (7 h e a d s) = P (9 h e a d s) \Rightarrow n C_{7} = n C_{9} \Rightarrow n = 16 ∴ P (3 h e a d s) = 16 C_{3} {(\frac{1}{2})}^{16} = \frac{35}{2^{12}} .$

Question 13

A five digit number is formed with digits 0. 1. 2. 3. 4 without repetition. A number is selected at random, then the probability that it is divisible by 4 is

\frac{1}{3}

\frac{5}{16}

\frac{1}{4}

\frac{4}{15}

SOLUTION

Solution : B

The number formed is divisible by 4 if the last two digits are 04,40,34,32,20,12.
Therefore total number of favorable ways = 3! + 3! +4 + 4+ 3! + 4 = 30 (This is the sum of number of ways in which the first two digits can be formed)
Total numbers that can be formed = 5! – 4! = 96 ( Number of ways of arranging 5 digits - Number of ways in which zero comes as the first digit)
Therefore required probability $= \frac{30}{96} = \frac{5}{16} .$

Question 14

On a toss of two dice, A throws a total of 5. Then the probability that he will throw another 5 before he throws 7 is

\frac{2}{45}

\frac{2}{5}

\frac{1}{81}

\frac{1}{9}

SOLUTION

Solution : B

let P(A) be the probability of throwing total of 5 and P(B) the probability of throwing total 7.
$∴ P (A) = \frac{4}{36} = \frac{1}{9}, P (B) = \frac{6}{36} = \frac{1}{6}, P (A o r B) = \frac{5}{18} a n d P (A a n d B) = 0 P (n e i t h e r A n o r B) = \frac{13}{18}, ∴ P (5 b e f o r e 7) = \frac{1}{9} + \frac{13}{18} . \frac{1}{9} + \frac{13}{18} . \frac{13}{18} . \frac{1}{9} + . . . . \infty = \frac{1}{9} [\frac{1}{1 - \frac{13}{18}}] = \frac{2}{5} .$

Question 15

A bag contains 'a' white and 'b' black balls. Two players A and B alternately draw a ball from the bag, replacing the ball each time after the draw. A begins the game. If the probability of A winning ( that is
drawing a white ball) is twice the probability of B winning, then the ratio a : b is equal to

A. 1 : 2

B. 2 : 1

C. 1 : 1

D. 1 : 3

SOLUTION

Solution : C

Let the event when a white ball is draw be W and for black ball let it be B. So A wins when we get sequence of the form
W or WBW or WBBBW or WBBBBBW......
Probability of getting W is a/a+b . So we get
$P (A) = \frac{a}{a + b} + {(\frac{b}{a + b})}^{2} \frac{a}{a + b} + {(\frac{b}{a + b})}^{4} \frac{a}{a + b} + . . . \infty = \frac{a + b}{a + 2 b} S i m i l a r l y w e g e t P (B) = \frac{b}{a + b} . \frac{a}{a + b} + {(\frac{b}{a + b})}^{3} \frac{a}{a + b} + . . . \infty = \frac{b}{a + 2 b} P (A) = 2 P (B) \Rightarrow a + b = 2 b \Rightarrow a = b .$

Free Objective Test 01 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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