Free Probability 01 Practice Test - 12th Grade - Commerce

Question 1

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 2

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 3

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 4

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 5

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 6

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 7

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 8

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 9

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
$P (A) = \frac{a}{a + b} + {(\frac{b}{a + b})}^{2} \frac{a}{a + b} + {(\frac{b}{a + b})}^{4} \frac{a}{a + b} + . . . \infty = \frac{a + b}{a + 2 b} S i m i l a r l y w e g e t P (B) = \frac{b}{a + b} . \frac{a}{a + b} + {(\frac{b}{a + b})}^{3} \frac{a}{a + b} + . . . \infty = \frac{b}{a + 2 b} P (A) = 2 P (B) \Rightarrow a + b = 2 b \Rightarrow a = b .$

Question 10

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 $= \frac{55}{100} = \frac{11}{20} .$

Free Probability 01 Practice Test - 12th Grade - Commerce

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

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Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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Application of Derivatives 02
Relations and Functions 01
Definite Integrals and Areas 01