Free Probability 02 Practice Test - 12th Grade - Commerce

Question 1

A random variable X has the following probability distribution
$\begin{matrix} X & P (X = x) & X & P (X = x) 0 & λ & 5 & 11 λ 1 & 3 λ & 6 & 13 λ 2 & 5 λ & 7 & 15 λ 3 & 7 λ & 8 & 17 λ 4 & 9 λ \end{matrix}$
then, $λ$ is equal to

\frac{1}{81}

\frac{2}{81}

\frac{5}{81}

\frac{7}{81}

SOLUTION

Solution : A

$\sum_{x = 0}^{8} P (X) = 1 \Rightarrow P (0) + P (1) + P (2) + P (3) + P (4) + P (5) + P (6) + P (7) + P (8) = 1 = λ + 3 λ + 5 λ + 7 λ + 9 λ + 11 λ + 13 λ + 15 λ + 17 λ = 1 \Rightarrow \frac{9}{2} (λ + 17 λ) = 1 \Rightarrow 81 λ = 1 ∴ λ = \frac{1}{81}$

Question 2

A man is known to speak truth is 75% cases. If he throws an unbiased die and tells his friend that it is a six, then the probability that it is actually a six, is

\frac{1}{6}

\frac{1}{8}

\frac{3}{4}

\frac{3}{8}

SOLUTION

Solution : D

Let E denote the event that a six occurs and A be the event theat the man reports that it is a six. We have $P (E) = \frac{1}{6}, P (E^{'}) = \frac{5}{6}, P (\frac{A}{E}) = \frac{3}{4}$ and $P (\frac{A}{E^{'}}) = \frac{1}{4}$ . By Baye’s theorem
$P (\frac{E}{A}) = \frac{P (E) . P (\frac{A}{E})}{P (E) . P (\frac{A}{E}) + P (E^{'}) . P (\frac{A}{E^{'}})} = \frac{(\frac{1}{6}) (\frac{3}{4})}{(\frac{1}{6}) (\frac{3}{4}) + (\frac{5}{6}) (\frac{1}{4})} = \frac{3}{8}$

Question 3

Seven coupons are selected at random one at a time with replacement from 15 coupons numbered 1 to 15. The probability that the largest number appearing on a selected coupon is 9, is

{(\frac{9}{16})}^{6}

{(\frac{8}{15})}^{7}

{(\frac{3}{5})}^{7}

{(\frac{3}{5})}^{7}

{(\frac{8}{15})}^{7}

SOLUTION

Solution : D

Each coupon can be selected in 15 ways. The total number of ways of choosing 7 copouns is $15^{7}$ . If largest number is 9, then the selected numbers have to be from 1 to 9 excluding those consisting of only 1 to 8.
Probability desired is $\frac{9^{7} - 8^{7}}{15^{7}}$
$= {(\frac{3}{5})}^{7} - {(\frac{8}{15})}^{7}$

Question 4

Team A plays with 5 other teams exactly once. Assuming that for each match the probabilities of a win, draw and loss are equal, then

A. The probability that A wins and loses equal number of matches is

\frac{34}{81}

B. The probability that A wins and loses equal number of matches is

\frac{17}{81}

C. The probability that A wins more number of matches than it loses is

\frac{17}{81}

D. The probability that A loses more number of matches than it wins is

\frac{16}{81}

SOLUTION

Solution : B

Probability of equal number of W and L is the probability of following
(0)W,(0)L +(1)W,(1)L+(2)W, (2)L
$= {(\frac{1}{3})}^{5} +^{5} C_{1} .^{4} C_{1} {(\frac{1}{3})}^{5} +^{5} C_{2} .^{3} C_{2} {(\frac{1}{3})}^{5} = \frac{17}{81}$

Question 5

The probabilities of three mutually exclusive events A, B, C are given by $\frac{2}{3}, \frac{1}{4}, a n d \frac{1}{6}$ respectively. This statement

A. Is true

B. is false

C. nothing can be said

D. could be either

SOLUTION

Solution : B

Since the events A,B,C are mutually exclusive, we have
$P (A ⋃ B ⋃ C) = \frac{2}{3} + \frac{1}{4} + \frac{1}{6} = \frac{13}{12} > 0$
Which is not possible , hence the statement is false.

Question 6

Two events A and B have the probabilities 0.25 and 0.5 respectively. The probability that both A and B simultaneously is 0.12. then the probability that neither A and nor B occurs is

A. 0.13

B. 0.38

C. 0.63

D. 0.37

SOLUTION

Solution : D

$P (A ⋃ B) = P (A) + P (B) - P (A ⋂ B) = 0.25 + 0.5 - 0.12 = 0.63$
Therefore required probability = 1 - P(A intersection B) = 0.37

Question 7

If $\frac{(1 + 4 p)}{4}, \frac{(1 - p)}{2}, and \frac{(1 - 2 p)}{2}$ are the probabilities of three mutually exclusive events , then the value of p is

\frac{1}{2}

\frac{1}{3}

\frac{1}{4}

\frac{1}{5}

SOLUTION

Solution : A

$\frac{1 + 4 p}{4}, \frac{1 - p}{2}, \frac{1 - 2 p}{2}$ are the probabilities of three mutually exclusive events.

$∴ 0 \leq \frac{1 + 4 p}{p} \leq 1, 0 \leq \frac{1 - p}{2} \leq 1$

$0 \leq \frac{1 - 2 p}{2} \leq 1 a n d 0 \leq (\frac{1 + 4 p}{4}) + (\frac{1 - p}{2}) + \frac{1 - 2 p}{2} \leq 1$

$∴ \frac{- 1}{4} \leq p \leq \frac{3}{4} 1 - \leq p \leq 1, \frac{1}{2} \leq p \leq \frac{1}{2} a n d p \leq \frac{5}{2}$

$\Rightarrow p = \frac{1}{2}$

Question 8

A father has three children with at least one boy. The probability that he has two boys and one girl is

\frac{1}{4}

\frac{3}{7}

\frac{1}{3}

\frac{2}{5}

SOLUTION

Solution : B

Let A be the event that father has at least one boy and B be the event that he has 2boys and one girl.
$P (A) = P (1 b o y, 2 g i r l) + P (2 b o y, 1 g i r l) + P (3 b o y, n o g i r l) = \frac{3 C_{1} + 3 C_{2} + 3 C_{3}}{2^{3}} = \frac{7}{8}$
$P (A ⋂ B) = P (2 b o y, 1 g i r l) = \frac{3}{8}$
$H e n c e P (\frac{B}{A}) = \frac{P (A ⋂ B)}{P (A)} = \frac{3}{7} .$

Question 9

A fair coin is tossed n times and let X denotes the number of heads obtained. If P(X = 4), P(X = 5), and P(X = 6) are in A.P., then n is equal to

A. Only 7

B. 14

C. 7 or 14

D. 8

SOLUTION

Solution : B and C

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.
This means $n C_{4} {(\frac{1}{2})}^{n}, n C_{5} {(\frac{1}{2})}^{n}, n C_{6} {(\frac{1}{2})}^{n} a r e A . P .$
$\Rightarrow 2 \times n C_{5} = n C_{4} + n C_{6}$
$\Rightarrow 2 \times \frac{n!}{(n - 5)! 5!} = \frac{n!}{(n - 4)! 4!} + \frac{n!}{(n - 6)!}!$
$\Rightarrow 6 \times 5 + (n - 4) (n - 5) = 2 \times 6 (n - 4)$
$\Rightarrow n^{2} - 9 n + 50 = 2 (6 n - 24)$
$\Rightarrow n^{2} - 21 n + 98 = 0$
$\Rightarrow n = 7 o r n = 14.$

Question 10

One hundred identical coins, each with probability p of showing up heads are tossed once. If 0 < p < 1 and the probability of heads showing 50 coins is equal to that head showing 51 coins, then the value of p is

\frac{1}{2}

\frac{49}{101}

\frac{50}{101}

\frac{51}{101}

SOLUTION

Solution : D

Let P1 and P2 be the respectively be the probability of heads showing 50 coins and head showing 51 coins $P_{1} = 100 C_{50} p^{50} (1 - p)^{50} a n d P_{2} = 100 C_{51} p^{51} (1 - p)^{49}$
$G i v e n P_{1} = P_{2}$
$\Rightarrow 100 C_{50} p^{50} (1 - p)^{50} = 100 C_{51} p^{51} (1 - p)^{49}$
$\Rightarrow \frac{1 - p}{p} = \frac{50}{51} \Rightarrow p = \frac{51}{101} .$

Free Probability 02 Practice Test - 12th Grade - Commerce

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

Related ExamsDefinite Integrals and Areas 02Differential Equations 03Probability 03

Related Exams
Definite Integrals and Areas 02
Differential Equations 03
Probability 03