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 = $\displaystyle \frac{1}{2} \times \frac{3}{8} + \frac{1}{2} \times \frac{2}{5} = \frac{3}{16} + \frac{1}{5} = \frac{15 + 16}{80} = \frac{31}{80}$
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%.
If two balls are drawn out of 6 Red + 8 Green balls = _{14}c_{2} = $\displaystyle \frac{14 \times 13 \times 12!}{12! \ 2!}$ = 91 ways.
Two Green balls can be drawn in _{8}c_{2} ways = $\displaystyle \frac{8 \times 7 \times 6!}{6! \ 2!}$ = 28.
Probability of drawing 2 Green balls = $\displaystyle \frac{28}{91} = \frac{4}{13}$
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 â€“ 25 = 176.
Probability of not divisible by 8 = $\displaystyle \frac{176}{201}$
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 = $\displaystyle \frac{2}{7}$
Exhaustive number of cases = 6^{2} = 36.
Favourable cases are (3,6), (6,3), (4,5), (5,4).
i.e. favourable ways = 4.
Hence, probability = $\displaystyle \frac{4}{36} = \frac{1}{9}$
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 => $\displaystyle 1 - \frac{1}{4!} = 1 - \frac{1}{24} = \frac{23}{24}$.
Let A = the event of drawing a spade
And B = event of drawing an ace
A and B are not mutually exclusive
Aâˆ©B = the event of drawing the ace of spades
$$ P(A) = \frac{13}{52}, \ P(B) = \frac{4}{52}, \ P(Aâˆ©B) = \frac{1}{52} $$P(AUB) = P(A) + P(B) â€“ P(Aâˆ©B)
$$ \frac{13}{52} + \frac{4}{52} â€“ \frac{1}{52} = \frac{16}{52} = \frac{4}{13} $$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.
Here P(A) = 20%, P(B) = 16%, P(C) = 14%
P(Aâˆ©B) = 8%, P(Aâˆ©C) = 5%, P(Bâˆ©C) = 4% and P(Aâˆ©Bâˆ©C) = 2%
P(AUBUC) = the probability that the person reads A or B or C
= P(A) + P(B) + P(C) - P(Aâˆ©B) - P(Aâˆ©C) - P(Bâˆ©C) + P(Aâˆ©Bâˆ©C)
= 20% + 16% + 14% - 8% - 5% - 4% + 2% = 35%
Hence, the probability that he reads none of the papers
P(AUBUC) = 1 - P(AUBUC)
$$ 1 - 35\% = 1 - \frac{35}{100} = \frac{65}{100} = 65\% $$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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