Probability Questions - 1

Bag A contains 3 Blue balls and 5 White balls. Bag B contains 4 Blue and 6 White balls. One of the bags is chooses at random and a ball is drawn from it find the probability that the ball is blue ?
  • $\displaystyle \frac{21}{50}$
  • $\displaystyle \frac{31}{50}$
  • $\displaystyle \frac{21}{80}$
  • $\displaystyle \frac{31}{80}$
  • $\displaystyle \frac{24}{37}$
Explanation   

Probability of choosing Bag A = $\displaystyle \frac{1}{2}$ , Bag B = $\displaystyle \frac{1}{2}$

Choosing Blue ball in Bag A = $\displaystyle \frac{3}{8}$ , Bag B = $\displaystyle \frac{4}{10}$ or $\displaystyle \frac{2}{5}$

Probability = $\displaystyle \frac{1}{2} \times \frac{3}{8} + \frac{1}{2} \times \frac{2}{5} = \frac{3}{16} + \frac{1}{5} = \frac{15 + 16}{80} = \frac{31}{80}$

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A speaks the truth in 90% of the cases; B speaks the truth in 80% of the cases, what is the probability that they tell the truth at the same time ?
  • 60%
  • 75%
  • 72%
  • 63%
  • 65%
Explanation   

Speaking truth by A = 90% = $\displaystyle \frac{9}{10}$, By B = 80% = $\displaystyle \frac{8}{10}$ or $\displaystyle \frac{4}{5}$

Tell the truth at the same time = $\displaystyle \frac{9}{10} \times \frac{4}{5} = \frac{18}{25} \times 100$ = 72%.

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Find the chance of drawing two Green balls in succession from a bag containing 6 Red 8 Green balls.
  • $\displaystyle \frac{4}{13}$
  • $\displaystyle \frac{6}{13}$
  • $\displaystyle \frac{7}{13}$
  • $\displaystyle \frac{8}{13}$
  • $\displaystyle \frac{4}{7}$
Explanation   

If two balls are drawn out of 6 Red + 8 Green balls = 14c2 = $\displaystyle \frac{14 \times 13 \times 12!}{12! \ 2!}$ = 91 ways.

Two Green balls can be drawn in 8c2 ways = $\displaystyle \frac{8 \times 7 \times 6!}{6! \ 2!}$ = 28.

Probability of drawing 2 Green balls = $\displaystyle \frac{28}{91} = \frac{4}{13}$

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Out of all the three digit integers between 1 to 300, a three digit number has to be selected at random, what is the probability that the selected number is not divisible by 8 ?
  • $\displaystyle \frac{182}{201}$
  • $\displaystyle \frac{177}{201}$
  • $\displaystyle \frac{176}{201}$
  • $\displaystyle \frac{126}{201}$
  • $\displaystyle \frac{132}{201}$
Explanation   

Number of three digit numbers from 1 to 300 = 201.

Number of three digit numbers divisible by 8 up to 300 = 25.

Not divisible by 8 = 201 – 25 = 176.

Probability of not divisible by 8 = $\displaystyle \frac{176}{201}$

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Probability of a leap year selected randomly has 53 Sundays.
  • $\displaystyle \frac{1}{7}$
  • $\displaystyle \frac{1}{3}$
  • $\displaystyle \frac{1}{8}$
  • $\displaystyle \frac{2}{7}$
  • None
Explanation   

A leap year contains 366 days = 52 complete weeks and 2 extra days.

To get 53 Sundays in the extra two days one of the day must be Sunday.

Probability of get 53 Sundays = $\displaystyle \frac{2}{7}$

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What is the probability of obtaining a total score of 9 in a single throw with two dice?
  • $\displaystyle \frac{1}{9}$
  • $\displaystyle \frac{2}{9}$
  • $\displaystyle \frac{1}{7}$
  • $\displaystyle \frac{1}{8}$
  • $\displaystyle \frac{2}{7}$
Explanation   

Exhaustive number of cases = 62 = 36.

Favourable cases are (3,6), (6,3), (4,5), (5,4).

i.e. favourable ways = 4.

Hence, probability = $\displaystyle \frac{4}{36} = \frac{1}{9}$

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There are 4 letters and 4 addressed envelopes. If the letters are in the envelopes at random what is the probability that all the letters are not placed in the right envelopes?
  • $\displaystyle \frac{24}{24}$
  • $\displaystyle \frac{23}{24}$
  • $\displaystyle \frac{22}{24}$
  • $\displaystyle \frac{23}{23}$
  • $\displaystyle \frac{22}{23}$
Explanation   

The total number of ways of placing 4 letters in 4 envelopes = 4!

All the letters can be placed correctly in only one way.

The probability that letters are placed in right envelopes = $\displaystyle \frac{1}{4!}$

Hence the probability that letters are not placed in right envelopes => $\displaystyle 1 - \frac{1}{4!} = 1 - \frac{1}{24} = \frac{23}{24}$.

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A card is drawn from a well shuffled pack of playing cards. What is the probability that is either a spade or an ace?
  • $\displaystyle \frac{\ 2}{13}$
  • $\displaystyle \frac{\ 5}{13}$
  • $\displaystyle \frac{\ 3}{13}$
  • $\displaystyle \frac{\ 4}{13}$
  • $\displaystyle \frac{\ 1}{13}$
Explanation   

Let A = the event of drawing a spade

And B = event of drawing an ace

A and B are not mutually exclusive

A∩B = the event of drawing the ace of spades

$$ P(A) = \frac{13}{52}, \ P(B) = \frac{4}{52}, \ P(A∩B) = \frac{1}{52} $$

P(AUB) = P(A) + P(B) – P(A∩B)

$$ \frac{13}{52} + \frac{4}{52} – \frac{1}{52} = \frac{16}{52} = \frac{4}{13} $$
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Three newspapers A,B,C are published in a city and a survey of readers indicates the following:

20% read A, 16% read B, 14% read C

8% read both A and B, 5% read both A and C

4% read both B and C, 2% read all the three

For a person chosen at random, find the probability that he read none of the papers.

  • 55%
  • 75%
  • 65%
  • 45%
  • 35%
Explanation   

Here P(A) = 20%, P(B) = 16%, P(C) = 14%

P(A∩B) = 8%, P(A∩C) = 5%, P(B∩C) = 4% and P(A∩B∩C) = 2%

P(AUBUC) = the probability that the person reads A or B or C

= P(A) + P(B) + P(C) - P(A∩B) - P(A∩C) - P(B∩C) + P(A∩B∩C)

= 20% + 16% + 14% - 8% - 5% - 4% + 2% = 35%

Hence, the probability that he reads none of the papers

P(AUBUC) = 1 - P(AUBUC)

$$ 1 - 35\% = 1 - \frac{35}{100} = \frac{65}{100} = 65\% $$
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There are 2 bags one of which contains 5 red balls and 7 white balls and the other 3 red balls and 12 white balls and a ball is to be drawn from one or other of the two bags. Find the chance of drawing a red ball?
  • $\displaystyle \frac{\ 35}{120}$
  • $\displaystyle \frac{\ 37}{120}$
  • $\displaystyle \frac{\ 30}{120}$
  • $\displaystyle \frac{\ 34}{120}$
  • $\displaystyle \frac{\ 31}{120}$
Explanation   

The chance of choosing the first bag is $\displaystyle \frac{1}{2}$, and if the first bag be chosen the chance of drawing a red ball from it is $\displaystyle \frac{5}{12}$, hence the chance of drawing a red ball from the first bag is => $\displaystyle \frac{1}{2} \times \frac{5}{12} = \frac{5}{24}$

Similarly the chance of drawing a red ball from the second bag is => $\displaystyle \frac{1}{2} \times \frac{3}{15} = \frac{1}{10}$

Hence as these events are mutually exclusive, the chance required is => $\displaystyle \frac{5}{24} + \frac{1}{10} = \frac{37}{120}$

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