Probability II

1) When a die is rolled, what is the probability of getting a number which has odd numbers?

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Sample space S = {1, 2, 3, 4, 5, 6}
Possible outcomes = {1, 3, 5}
Probability = \({n(e) \over {n(s)}}\) =  \({1 \over {2}}\) Hence (a).

2) If a 3 digit number is made out using digits 1, 2, 3, 4, 6 and 8 only, then what is the probability that the number is divisible by 2?

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Total outcome = 6 3
Possible outcome =  4 = 144
Probability = \({n(e) \over {n(s)}}\) =  \({144 \over {216}}\) =  \({2 \over {3}}\) Hence (b).

3) A fair coin is randomly tossed 11 times. What is the probability of getting 7 heads and 3 tails?

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4) Two friends A and B attended the interview. The probability of getting selected A is \({1 \over {5}}\) and probability of getting selected B is \({3 \over {7}}\). Find the probability that both of them are selected

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Probability of getting A selected  =  \({1 \over {5}}\)

Probabilit y of b getting selected  =  \({3 \over {7}}\)

Probability that both are selected =  \({1 \over {5}}*{3 \over {7}}\) =  \({3 \over {35}}\) Hence (c).

5) A bag contains 7 pink, 9 blue and 6 white toys. If 3 toys are picked at random, what is the probability that 1toy must be pink?

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Probability of picking 3 toys such that 1 toy from pink and other two from blue and white toy =  \({7c* 15c \over {22c}}\) =  \({21 \over {44}}\) Hence (b).

6) In a set of first 60 natural numbers, what is the probability of getting multiples of 4 and sum of the digits of individual numbers should be a prime number?

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Total outcome = {1, 2, 3, 4,…..60}

Number of favourable outcome = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60}

On the above only 12, 16, 20, 32, 48, 52 and 56 are multiples of 4 and also the sum of digits of individual number is prime number.

Probability = \({7 \over {60}}\) Hence (c).

7) All the face cards are removed from a deck of 52 cards and then shuffled. A card is drawn from the remaining cards. Find the probability of getting 8 or 9 of red colour.

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Total number of cards = 52
All face cards are removed, then remaining cards = 52 – 12 = 40
Number of ‘8’ or ‘9’ of red colour is one each from  hearts and diamonds = 4

Probability =  \({4 \over {40}}\) =  \({1 \over {10}}\) Hence (b).

8) Of the 900 participants in a Conference hall, 630 are females. \({3 \over {7}}th\)of the female and\({1 \over {3}}rd\) of the male participants are less than twenty five years old. If one of the participants is randomly selected to receive a prize, what is the probability that the person selected will be less than twenty five years old?

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Total participants =9 00
Female participants = 6 3 0
Male participants =900- 6 30 =27 0
Participants less than twen ty five years
\({3 \over {7}}\) ( 6 3 0) + \({1 \over {3}}\) (270) = 2 70 +90 =36 0
Probability that the person selecte d will be less than twenty five  years old = \({360 \over {900}}\) = \({2 \over {5}}\)
Hence (c).

9)

If the odds in favour of P solving a question are 9 to 5 and odds against Q solving same question are 3 to 7, find the probability of P and Q solving the question.

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P solving the question à Number of favourable outcomes = 9 Number of unfavourable outcomes = 5 P(P) = \({Number of favourable outcomes \over {Number of favourable outcomes+Number of unfavourable outcomes}}\) = \({9 \over {9+ 5}}\) = \({9 \over {14}}\) Q solving the question à Number of favourable outcomes = 7 Number of unfavourable outcomes = 10 P(Q) = \({Number of unfavourable outcomes \over {Number of favourable outcomes+Number of unfavourable outcomes}}\) = \({7 \over {7+ 3}}\) = \({7 \over {10}}\) Probability of P and Q solving the question = P(P) * P(Q) = \({9 \over {14}}\) * \({7 \over {10}}\) = \({9 \over {20}}\)Hence (d).

10)

When 2 dice are thrown simultaneously, what is the probability that the sum showing on the
Upper face is less than 9?

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Possibility of getting a sum of 12 is (6, 6) à 1 Possibility of getting a sum of 11 is (5, 6), (6, 5) à 2 Possibility of getting a sum of 10 is (4, 6), (5, 5), (6, 4)à 3 Possibility of getting a sum of 9 is (3, 6), (4, 5)à 2 Number of favourable outcome = 36 – 8 = 28 Probability of getting sum of two numbers less than 9 = \({7 \over {9}}\)Hence (a).

11) 7 coins are tossed simultaneously. What is the probability of getting atmost 3 tails?

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Number of favourable  outcomes (no tail (or) 1 tail (or) 2 tails (or) 3 tails) =  7c 0 + 7c 1 + 7c 2 + 7c 3
= 1 + 7 + 21 + 35 = 64
Total outcomes when 7 coins are tossed = 2 7 = 128
Probability = \({64 \over {128}}\) =  \({1 \over {2}}\) Hence (d).

12) In a box holding 32 mangoes, 37.5% turned out to be rotten. If 4 mangoes are taken out from the box, what is the probability that at least one of the four mangoes picked up is good?

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Total outcomes = 32c 4 = 35960

Rotten mangoes  = 37.5% of 32 = 12

Number of ways of selecting 4 mangoes out of 12 = 12c 4 = 495

Number of good mangoes to be selected =  35960 – 495 = 35465

Probability = \({35465 \over {35960}}\) =  \({7093 \over {7192}}\) Hence (c).

13) Four cards are drawn progressively, without replacement from a pack of 52 well-shuffled cards. What is the probability that the first, third and fourth cards are king and the second card drawn is another card?

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Probability f or the first card to be a king =  \({4 \over {52}}\)

Probability for the second card to be a different card = \({48 \over {51}}\)

Probability for the third card to be a king =  \({3 \over {50}}\)

Probability for the fourth card to be a king =  \({2 \over {49}}\)

Probability =  \({4 \over {52}}\) *  \({48 \over {51}}\) *  \({3 \over {50}}\) *  \({2 \over {49}}\) =  \({48 \over {270725}}\) .  Hence (a).

14)

What is the probability that 6 digit numbers can be formed from using only the digits 3, 4, 5, 6, 7 and 8(without repetition), when the digits at the unit place must be greater than that in the ten’s place?

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à If the digit in the unit place is 8, the remaining 5 digits can be placed in 1* 2* 3 * 4* 5 * 1 = 5! = 120 ways àIf the digit in the unit place is 7, then the ten’s digit can be 3, 4, 5 or 6. The remaining 5 digits can be placed in 1 * 2 * 3 * 4 * 4 * 1 = 4 * 4! = 96 ways à If the digit in the unit place is 6, then the ten’s digit can be 3, 4, or 5. The remaining 5 digits can be placed in 1 * 2 * 3 * 4 * 3 * 1 = 3 * 4! = 72 ways à If the digit in the unit place is 5, then the ten’s digit can be 3, or 4.The remaining 5 digits can be placed in 1 * 2 * 3 * 4 * 2 * 1 = 2 * 4! = 48 ways à If the digit in the unit place is 4, then the ten’s digit can be 3. The remaining 5 digits can be placed in 1 * 2 * 3 * 4 * 1 * 1 = 1 * 4! = 24 ways No possible arrangement with 3 in unit place. Total number of ways = (120 + 96 + 72 + 48 + 24) = 360. Probability = \({360 \over {720}}\) = \({1 \over {2}}.\)Hence (b).

15) Three marbles are drawn at random from 7 green, 6 brown and 5 pink marbles. What is the probability of getting at least 1 pink and 1 green marble?

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Pr obability of selecting  1 green, 1 blue, 1 pink marble =  \({7c \over {18c}}\) *  \({6c \over {17c}}\) *  \({5c \over {16c}}\) =  \({7 \over {18}}\) *  \({6 \over {17}}\) *  \({5 \over {16}}\)

(or)

Probability of selecting 1 green, 1 green, 1 pink marble = \({7c \over {18c}}\) *  \({6c \over {17c}}\) *  \({5c \over {16c}}\) =  \({7 \over {18}}\) *  \({6 \over {17}}\) *  \({5 \over {16}}\)

(or)

Probability of selecting at least 1 green, 1 blue, 1 pink marble =  \({5c \over {18c}}\) *  \({4c \over {17c}}\) *  \({7c \over {16c}}\) =  \({5 \over {18}}\) *  \({4 \over {17}}\) *  \({7 \over {16}}\)

Probability = \({210+210 140 \over {18*17*16}}\) =  \({35 \over {306}}\) .  Hence (d).