Question

A bag contain 3 White,4 red and 5 black balls .Two balls are drawn one by one without replacement. what is the probability that at least one ball is black?

Answer 1

white balls=3
red balls =4
black balls=5
Total balls=12

P(atleast 1 ball is black)= 1-P(no ball is black)
$= 1 - \frac{c(7 \: 2)}{c(12 \: 2)} \\ = 1 - \frac{7 \times 6}{12 \times 11} \\ = 1 - \frac{7}{22} \\ = \frac{22 - 7}{22} \\ = \frac{15}{22}$

Answer 2

Hello!

A bag contains 3 white 4 red and 5 black balls two balls are drawn one by one without replacement what is the probability that at least one ball is black

We have the following information:

S (sample space) = 12 balls

W (white ball event) = 3 balls

R (red ball event) = 4 balls

B (black ball event) = 5 balls

PB (probability that at least one ball is black) =?

Solving:

1) First, calculate the probability that none of two ball are black (White and Red):

* The first withdrawal of the black ball, we have:

$P(WR)_1 = \dfrac{W+R}{S} \to P(WR)_1 = \dfrac{3+4}{12} \to \boxed{P(WR)_1 = \dfrac{7}{12}}$

** The second withdrawal of the black ball, if a black ball has already been retired (without replacement), we have:

$P(WR)_2 = \dfrac{(W+R)-1}{S-1} \to P(WR)_2 = \dfrac{(3+4)-1}{12-1} \to P(WR)_2 = \dfrac{7-1}{11} \to \boxed{P(WR)_2 = \dfrac{6}{11}}$

2) Now, calculate the probability at least one ball is black, let's see:

$PB = 1 - P(WR)_1*P(WR)_2$

$PB = 1 - \dfrac{7}{12} * \dfrac{6}{11}$

$PB = 1 - \dfrac{42}{132}$

$PB = \dfrac{132-42}{132}$

$PB = \dfrac{90}{132}$

simplify by 6  

$PB = \dfrac{90}{132}\frac{\div6}{\div6}$

$\boxed{\boxed{PB = \dfrac{15}{22}}}\end{array}}\qquad\checkmark$

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Another way to do it (by simple combination), let's see:  

PB = (just a black ball) 1 - (no black ball) simple combinations

$PB = 1 - \dfrac{C_{7,2}}{C_{12,2}}$

$PB = 1 - \dfrac{\dfrac{7!}{2!(7-2)!} }{\dfrac{12!}{2!(12-2)!}}$

$PB = 1 - \dfrac{\dfrac{7!}{2!5!} }{\dfrac{12!}{2!10!}}$

$PB = 1 - \dfrac{\dfrac{7*6*\diagup\!\!\!\!5!}{2!\diagup\!\!\!\!5!} }{\dfrac{12*11*\diagup\!\!\!\!\!10!}{2!\diagup\!\!\!\!\!10!}}$

$PB = 1 - \dfrac{\dfrac{7*6}{2*1}}{\dfrac{12*11}{2*1}}$

$PB = 1 - \dfrac{\dfrac{42}{2}}{\dfrac{132}{2}}$

$PB = 1 - \dfrac{21}{66}$

simplify by 3  

$PB = 1 - \dfrac{21}{66}\frac{\div3}{\div3}$

$PB = 1 - \dfrac{7}{22}$

$PB = \dfrac{22-7}{22}$

$\boxed{\boxed{PB = \dfrac{15}{22}}}\end{array}}\qquad\checkmark$

Answer:  

15/22

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I Hope this helps, greetings ... Dexteright02! =)