Free Probability 03 Practice Test - 12th Grade - Commerce

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

Three six faced fair dice are rolled together. The probability that the sum of the numbers appearing on the dice is 8, is

\frac{7}{72}

\frac{7}{54}

\frac{4}{27}

\frac{7}{64}

SOLUTION

Solution : A

Getting sum 8 when three dice are rolled = E
The possibilities are (1,1,6) , (2,2,4), (3,3,2), (1,2,5), (1,3,4)
The number of favorable cases for E = 3 + 3 + 3 + 6 + 6 = 21
The total number of cases = $6^{3}$ = 216
Required probability P(E) $= \frac{21}{216} = \frac{7}{72} .$

Question 7

In a multiple choice question there are four alternative answers of which one or more than one is or are correct. A candidate will get marks on the question only if he ticks all correct answers. The candidate decides to tick answers at random. If he is allowed up to three chances to answer the question, the probability that he will get marks on it is given by

1 - {(\frac{14}{15})}^{3}

{(\frac{1}{15})}^{3}

\frac{1}{5}

\frac{14}{15}

SOLUTION

Solution : C

The total number of ways of answering one or more alternatives of 4 is $4 C_{1} + 4 C_{2} + 4 C_{3} + 4 C_{4} = 15,$
Out of these 15 combinations, only one combination is correct. The probability of answering the alternaive correctly at the first trial is $\frac{1}{15},$ that at the second trial is $(\frac{14}{15} . \frac{1}{14}) = \frac{1}{15,}$ and that at the third trail is $\frac{14}{15} . \frac{13}{14} . \frac{1}{13} = \frac{1}{15},$
Therefore the probability that the andidate will get marks on the question if he allowed upto three chances, is $\frac{1}{15} + \frac{1}{15} + \frac{1}{15} = \frac{1}{5} .$

Question 8

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

\frac{32}{55}

\frac{30}{40}

\frac{32}{50}

\frac{3}{4}

SOLUTION

Solution : A

case(i): red ball from A to B, red ball from B to A ,then red ball from A
$P_{1} = \frac{6}{10} \times \frac{5}{11} \times \frac{6}{10}$
Case(ii): red ball from A to B, white ball from B to A, red ball from A
$P_{2} = \frac{6}{10} \times \frac{6}{11} \times \frac{5}{10}$
Case(iii): white ball from A to B, red ball from B to A, red ball from A
$P_{3} = \frac{4}{10} \times \frac{4}{11} \times \frac{7}{10}$
Case(iv): white ball from A to B, white ball from B to A, red ball from A
$P_{4} = \frac{4}{10} \times \frac{7}{11} \times \frac{6}{10}$
Therefore required probability $= P_{1} + P_{2} + P_{3} + P_{4} = \frac{640}{1100} = \frac{32}{55}$

Question 9

From a bag containing 9 distinct white and 9 distinct black, 9 balls are drawn at random one by one, the drawn balls being replaced each time. The probability that at least four balls of each colour is in the draw, is

A. A little less than

\frac{1}{2}

B. a little greater than

\frac{1}{2}

\frac{1}{2}

\frac{1}{3}

SOLUTION

Solution : A

Out of 9 distinct black and 9 distinct white balls, probability of drawing a white ball $= \frac{1}{2}$
and drawing black ball is also $\frac{1}{2} .$ for at least 4 of each colour in 9 draws with replacement, there are two cases,
Case(i)
P(getting 5 white, 4 black) $= 9 C_{5} {(\frac{1}{2})}^{9}$
Cases (ii)
P(getting 4 white, 5 black) $= 9 C_{4} {(\frac{1}{2})}^{9}$
These are exclusive so
P(atleast 4 of each colour) $= 9 C_{5} {(\frac{1}{2})}^{9} + 9 C_{4} {(\frac{1}{2})}^{9} = \frac{63}{128}$ which is little less than $\frac{1}{2}$

Question 10

Let A, B , and C be three independent events with $P (A) = \frac{1}{3}, P (B) = \frac{1}{2}, a n d P (C) = \frac{1}{4} .$ . The probability of exactly 2 of these events occurring, is equal to

\frac{1}{4}

\frac{7}{24}

\frac{3}{4}

\frac{17}{24}

SOLUTION

Solution : A

P(exactly two of A, B and C) = P(AB) + P(BC) + P(CA) – 3 P(ABC)
Since A, B ,C are independent events,
Required probability = P(A)P(B) + P(B)P(C) +P(C)P(A) – 3 P(A)P(B)P(C)
$= \frac{1}{3} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{4} + \frac{1}{4} \times \frac{1}{3} - 3 \frac{1}{3} \times \frac{1}{2} \times \frac{1}{4} = \frac{1}{4.}$

Free Probability 03 Practice Test - 12th Grade - Commerce

Question 1

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Question 2

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Question 3

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Question 4

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Question 5

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Question 6

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Question 7

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Question 8

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Question 9

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Question 10

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