Probability - Study Mode

[#96] A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If four marbles are picked at random, what is the probability that none is blue?
Correct Answer

(B) $$frac{{33}}{{91}}$$

Explanation

Solution: Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. When four marbles are picked at random, then the probability that none is blue is $$eqalign{
& = frac{{{}^{12}{C_4}}}{{{}^{15}{C_4}}} cr
& = frac{{12 imes 11 imes 10 imes 9}}{{15 imes 14 imes 13 imes 12}} cr
& = frac{{11880}}{{32760}} cr
& = frac{{33}}{{91}} cr} $$

[#97] The probability that a number selected at random from the first 50 natural numbers is a composite number is -
Correct Answer

(B) $$frac{{17}}{{25}}$$

Explanation

Solution: The number of exhaustive events = $${{}^{50}{C_1}}$$ = 50 We have 15 primes from 1 to 50 Number of favourable cases are 34 Required probability = $$frac{{34}}{{50}}$$ = $$frac{{17}}{{25}}$$

[#98] If a card is drawn from a well shuffled pack of cards, the probability of drawing a spade or a king is -
Correct Answer

(D) $$frac{{4}}{{13}}$$

Explanation

Solution: $$Pleft( {S cup K}
ight) = $$ xa0 $$Pleft( S
ight) + $$xa0 $$Pleft( K
ight) - $$xa0 $$Pleft( {S cap K}
ight),$$ xa0 (where S denotes spade and K denotes king.) $$eqalign{
& Pleft( {S cup K}
ight) = frac{{13}}{{52}} + frac{4}{{52}} - frac{1}{{52}} cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,, = frac{4}{{13}} cr} $$

[#99] A bag contains five white and four red balls. Two balls are picked at random from the bag. What is the probability that they both are different color?
Correct Answer

(B) $$frac{{5}}{{9}}$$

Explanation

Solution: Two balls can be picked from nine balls in $${{}^9{C_2}}$$ ways. We select one white ball and one red ball from five white balls and four red balls. This can be done $${{}^5{C_1}}$$ . $${{}^4{C_1}}$$ ways. ∴ The required probability $$eqalign{
& = frac{{5 imes 4}}{{{}^9{C_2}}} cr
& = frac{{20}}{{36}} cr
& = frac{5}{9} cr} $$

[#100] A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that exactly three bulbs are good.
Correct Answer

(B) $$frac{{20}}{{63}}$$

Explanation

Solution: Required probability $$eqalign{
& = frac{{{}^5{C_3},.,{}^4{C_1}}}{{{}^9{C_4}}} cr
& = frac{{10 imes 4}}{{126}} cr
& = frac{{20}}{{63}} cr} $$