There are 7 red and 8 white balls in the box

Question:1

There are 7 red and 8 white balls in the box. We get 5 balls without returning. What is the probability that exactly 4 balls will be red ?

Solution:

Number of Red balls = 7

Number of white balls = 8

Probability that exactly 4 balls will be Red = 4 balls Red × One ball is White

$= \dfrac{{}^7C_4\times {}^8C_1}{{}^15C_5}$

$=\dfrac{ \dfrac{7!}{4!\times 3!}\times \dfrac{8!}{1!\times 7!}}{\dfrac{15!}{5!\times 10!}}$

$= \dfrac{\dfrac{7\times 6\times 5}{3\times 2 \times 1}\times 8}{\dfrac{15\times 14\times 13\times 12 \times 11 \times 10}{5\times 4 \times 3 \times 2 \times 1}}$

$= \dfrac{7 \times 5 \times 8}{7 \times 13 \times 3 \times 11}$

$= \dfrac{40}{429}$

= 0.0932

Question: 2

There are 7 blue balls and 5 black balls in a bag. Tony randomly draws 2 balls one by one from the bag without replacement. What is the probability that at least one black ball is drawn ?

Solution:

Number of Blue balls = 7

Number of black balls = 5

Let x is number of black balls

P(At lest one black balls) = Eaxactly one black ball + Exactly 2 black balls

P(x ≥ 1) = P(x = 1) + P(x = 2)

$= \dfrac{{}^5C_1\times {}^7C_1}{{}^{12}C_2} + \dfrac{{}^5C_2\times {}^7C_0}{{}^{12}C_2}$

Since, ${}^{12}C_2 = \dfrac{12!}{2! \times 10!}$

$= \dfrac{12 \times 11}{2\times 1}$

= 66

Now,

P(x ≥ 1) $= \dfrac{5 \times 7}{66} + \dfrac{\dfrac{5!}{2!\times 3!} \times 1}{66}$

$= \dfrac{35 + 10}{66}$

$= \dfrac{45}{66} = \dfrac{15}{22}$

= 0.6818

Therefore, the probability that at least one black ball is drawn when two balls are drawn without replacement from a bag containing 7 blue balls and 5 black balls is approximately 0.6818 or 68.18%.

Question: 3

There are 5 red balls, 6 white balls, 8 blue balls and 6 green balls in the lottery drum. We will randomly select 6 balls by sampling without replacement. What is the probability that all the selected balls are red ?

Solution:

Probability of all balls should be red are not possible because there is only 5 Red balls

Therefore, this event is imposible event

So, the probability of impossible event is = 0(Zero)

Question: 4

If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls. One ball is drawn from each box at random, what is the probability that two white and one black ball will be drawn ?

Solution:

In Box I, 3 white and 1 black

In box II, 2 white and 2 black

In box III, 1 white and 3 black balls

There are three ways of selection of 2 white ball and one black ball

Case I: white ball from box I, white ball from box II, black ball from box III

Probability = P(White ball)×P(White ball)×P(Black ball)

= 3/4 × 2/4 × 3/4 = 18/64

Case II: white ball from box I, black ball from box II, white ball from box III

Probability = P(White ball)×P(Black ball)×P(White ball)

= 3/4 × 2/4 × 1/4 = 6/64

Case III: Black ball from box I, White ball from box II, white ball from box III

Probability = P(Black ball)×P(White ballball)×P(White ball)

= 1/4 × 2/4 × 1/4 = 2/64

Thus, the probability that 2 white balls and 1 black ball = 18/64 + 6/64 + 2/64

= 26/64 = 0.40625

Question: 5

Consider two boxes, 1 containing 1 black ball and 2 white balls, the other, 3 black balls and 5 white balls. A box is selected at random, and a ball is drawn from the selected box. What is the probability that the ball is white?