Introduction Of Probability Theory

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If a ball is drawn at random from each of three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, then the probability that 2 white and 1 black balls will be drawn is:
1. $\frac{13}{32}$
2. $\frac{1}{4}$
3. $\frac{1}{32}$
4. $\frac{3}{16}$
A and B draw two cards, one after another, from a pack of well-shuffled pack of 52 cards. The probability that all the four cards drawn are of the same suit is:
1. None of these
2. $\frac{11}{85 \times 49}$
3. $\frac{13 \times 24}{17 \times 25 \times 49}$
4. $\frac{44}{85 \times 49}$
A and B are two events such that P(A) = 0.25 and P(B) = 0.50. The probability of both happening together is 0.14. The probability of both A and B not happening is:
1. 0.39
2. 0.25
3. 0.11
4. None of these
The probabilities of a student getting I, II and III division in an examination are $\frac{1}{10}$ , $\frac{3}{5}$ and $\frac{1}{4}$ respectively. The probability that the student fails in the examination is:
1. $\frac{197}{200}$
2. $\frac{27}{100}$
3. $\frac{83}{100}$
4. None of these
The conditional probability of an event E, given the occurrence of the event F is given by:
1. $P (E | F) = \frac{P (E \cap F)}{P (F)}, P (F) \neq 0$
2. $P (E | F) = \frac{P (E \cap F)}{P (E)}, P (F) \neq 0$
3. $P (E | F) = \frac{P (E \cap F)}{P (F)}, P (F) < 0$
4. $P (E | F) = \frac{P (E \cup F)}{P (F)}, P (F) \neq 0$
The conditional probability of an event E, given the occurrence of the event F lies between:
1. $0 < P (E | F) \leq 1$
2. $0 \leq P (E | F) < 1$
3. $0 \leq P (E | F) \leq 1$
4. $0 < P (E | F) < 1$
The conditional probability of the event E', given that F has occurred is given by:
1. $P (E' | F) = P (E | F)$
2. $P (E' | F) = - 1 + P (E | F)$
3. $P (E' | F) = 1$
4. $P (E' | F) = 1 - P (E | F)$
If E, F and G are events with $P (G) \neq 0$ then $P ((E \cup F) | G)$ is given by:
1. $P (E | G) + P (F | G) - P ((E \cap F) | F)$
2. $P (E | G) + P (G | F) - P ((E \cap F) | G)$
3. $P (E | G) + P (F | G) - P ((E \cap F) | G)$
4. $P (G | E) + P (F | G) - P ((E \cap F) | E)$
If E and F are events then $P (E \cap F)$ is equal to:
1. $P (E) P (F | E), P (E) \neq 0$
2. $P (E \cap F) P (F | E), P (E) \neq 0$
3. $P (E \cup F) P (F | E), P (E) \neq 0$
4. $P (E) P (E | F), P (E) \neq 0$
Two coins are tossed once, where E : tail appears on one coin, F : one coin shows head. Find $P (E / F)$ :
1. 1
2. 0.24
3. 0.23
4. 0.33
Two coins are tossed once ,where E :no tail appears , F : no head appears. Find $P (E / F)$ .
1. 0.22
2. 0.24
3. 0
4. 0.25
Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and $P (E \cap F) = 0.2$ , find $P (E / F)$ and $P (F / E)$ .
1. $P (E / F) = \frac{2}{4}, P (F / E) = \frac{1}{3}$
2. $P (E / F) = \frac{2}{3}, P (F / E) = \frac{1}{3}$
3. $P (E / F) = \frac{2}{5}, P (F / E) = \frac{1}{4}$
4. $P (E / F) = \frac{2}{3}, P (F / E) = \frac{1}{4}$
Compute P(A/B), if P(B) = 0.5 and $P (A \cap B) = 0.32$ .
1. $P (A / B) = \frac{16}{33}$
2. $P (A / B) = \frac{16}{25}$
3. $P (A / B) = \frac{16}{29}$
4. $P (A / B) = \frac{15}{27}$
If P (A) = 0.8, P (B) = 0.5 and P(B/A) = 0.4, find $P (A \cap B)$ .
1. 0.37
2. 0.29
3. 0.35
4. 0.32
If P (A) = 0.8, P (B) = 0.5 and P (B/A) = 0.4, find P (A/B).
1. 0.66
2. 0.62
3. 0.64
4. 0.68
If P (A) = 0.8, P (B) = 0.5 and P (B|A) = 0.4, find $P (A \cup B)$ .
1. 0.25
2. 0.95
3. 0.98
4. 1.00
Evaluate $P (A \cup B)$ , if $2 P (A) = P (B) = \frac{5}{13}$ and $P (A / B) = \frac{2}{5}$ .
1. $\frac{15}{26}$
2. $\frac{11}{27}$
3. $\frac{11}{26}$
4. $\frac{11}{29}$
If $P (A) = \frac{6}{11}$ , $P (B) = \frac{5}{11}$ and $P (A \cup B) = \frac{7}{11}$ , find P (A/B).
1. $\frac{4}{5}$
2. $\frac{3}{7}$
3. $\frac{7}{3}$
4. $\frac{5}{4}$
If $P (A) = \frac{6}{11}$ , $P (B) = \frac{5}{11}$ and $P (A \cup B) = \frac{7}{11}$ , find P (B/A).
1. $\frac{3}{5}$
2. $\frac{2}{3}$
3. $\frac{1}{3}$
4. $\frac{4}{5}$
A speaks truth in 75% cases and B speaks truth in 80% cases. Probability that they contradict each other in a statement is:
1. $\frac{7}{20}$
2. $\frac{13}{20}$
3. $\frac{3}{5}$
4. $\frac{2}{5}$
Out of 30 consecutive integers, 2 are chosen at random. The probability that their sum is odd is:
1. $\frac{14}{29}$
2. $\frac{16}{29}$
3. $\frac{15}{29}$
4. $\frac{10}{29}$
A coin is tossed three times. If events A and B are defined as A = Two heads come, B = Last should be head. Then, A and B are:
1. Independent
2. Dependent
3. Both independent and dependent
4. Mutually exclusive
If S is a sample space and $P (A) = \frac{1}{3} P (B)$ and $S = A \cup B$ , where A and B are two mutually exclusive events, then P (A) is:
1. $\frac{1}{4}$
2. $\frac{1}{2}$
3. $\frac{3}{4}$
4. $\frac{3}{8}$
If $P (A \cup B) = 0.8$ and $P (A \cap B) = 0.3$ , then $P (\bar{A}) + P (\bar{B})$ equals:
1. 0.3
2. 0.5
3. 0.7
4. 0.9
A bag contains 2 white and 3 black balls and another bag Y contains 4 white and 2 black balls. One bag is selected at random and a ball is drawn from it. Then the probability chosen to be white is:
1. $\frac{2}{15}$
2. $\frac{7}{15}$
3. $\frac{8}{15}$
4. $\frac{14}{15}$
If A and B are two independent events such that P (A) = 0.3, $P (A \cup B) = 0.5$ , then $P (A / B) - P (B / A)$ = :
1. $\frac{2}{7}$
2. $\frac{3}{35}$
3. $\frac{1}{70}$
4. $\frac{1}{7}$
Three persons A, B, and C fires a target in turn starting with A. Their probabilities of hitting the target are 0.4, 0.3 and 0.2 respectively. The probability of two hits is:
1. 0.024
2. 0.452
3. 0.336
4. 0.138
If two events are independent, then:
1. They must be mutually exclusive
2. The sum of their probabilities must be equal to 1
3. The sum of their probabilities must be equal to 0
4. None of these
A die is thrown and a card is selected at random from a deck of 52 playing cards. The probability of getting an even number on the die and a spade card is:
1. $\frac{1}{2}$
2. $\frac{1}{4}$
3. $\frac{1}{8}$
4. $\frac{3}{4}$
Let A and B be two events such that P (A) = 0.6, P (B) = 0.2 and $P (A / B) = 0.5$ . Then $P (\bar{A} / \bar{B})$ is:
1. $\frac{1}{10}$
2. $\frac{3}{10}$
3. $\frac{3}{8}$
4. $\frac{6}{7}$

Introduction Of Probability Theory

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