Question

(1) In a dark room, there are 10 black and 5 red balls in a box. If you pick 4 balls altogether, what is the probability at most 3 balls will be red?

(2) If you pick 7 cards together randomly from a deck of cards, what is the probability that there will be 5 black and 2 red cards?

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Answer 1

$\large\underline{\sf{Solution-1}}$

Box contains

• Number of Black balls = 10

• Number of Red balls = 5

So,

• Total number of balls in a box = 15

So,

• Number of ways in which four balls can be drawn from box is given by

$\rm :\longmapsto\:Total_{(outcomes)} = \:^{15} C_4$

So,

Probability of getting at most 3 red balls is

$\rm\:Probability_{(atmost \: 3 \: red \: balls)} = 1 - Probability_{(4 red balls)}$

$\rm \: Probability_{(atmost \: 3 \: red \: balls)} = 1 - \dfrac{\:^5 C_4}{\:^{15} C_4}$

$\rm \: Probability_{(atmost \: 3 \: red \: balls)} = 1 - \dfrac{5}{15} \times \dfrac{4}{14} \times \dfrac{3}{13} \times \dfrac{2}{12}$

$\rm \: Probability_{(atmost \: 3 \: red \: balls)} = 1 - \dfrac{1}{273}$

$\rm \: Probability_{(atmost \: 3 \: red \: balls)} = \dfrac{273 - 1}{273}$

$\rm \: Probability_{(atmost \: 3 \: red \: balls)} = \dfrac{272}{273}$

$\large\underline{\sf{Solution-2}}$

Total number of cards = 52

• Number of ways in which 7 cards can be picked randomly from pack of 52 cards is given by

$\rm :\longmapsto\:Total_{(outcomes)} = \:^{52} C_7$

Now,

We know,

• Number of black cards = 26

and

• Number of red cards = 26

So,

Number of ways in which 5 black balls and 2 red balls can be drawn from 26 black balls and 26 red balls is given by

$\rm \: Total_{(favourable \: outcomes)} = ^{26} C_5 \times ^{26} C_2$

So,

The probability that there will be 5 black and 2 red cards is

$\rm \: Probability_{(5 \: black \: and \: 2 \: red)} = \dfrac{^{26} C_5 \times ^{26} C_2}{^{52} C_7}$