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).

Possible outcomes = {1, 3, 5}

Probability = \({n(e) \over {n(s)}}\) = \({1 \over {2}}\) Hence (a).

Total outcome = 6 ^{3}

Possible outcome =__6 __ * __6 __ * __4__ = 144

Probability = \({n(e) \over {n(s)}}\) = \({144 \over {216}}\) = \({2 \over {3}}\) Hence (b).

Possible outcome =

Probability = \({n(e) \over {n(s)}}\) = \({144 \over {216}}\) = \({2 \over {3}}\) Hence (b).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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.

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).

When 2 dice are thrown simultaneously, what is the probability that the sum showing on the

Upper face is less than 9?

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).

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).

= 1 + 7 + 21 + 35 = 64

Total outcomes when 7 coins are tossed = 2

Probability = \({64 \over {128}}\) = \({1 \over {2}}\) Hence (d).

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).

Rotten mangoes = 37.5% of 32 = 12

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

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

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

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).

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).

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?

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).

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).

(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).