Consider an urn that contains $4$ blue and $7$ red balls

Question

I'm in need of some help with this problem. Consider an urn that contains $4$ blue and $7$ red balls. First one ball is selected, and then a second ball is selected without replacement.

(a) What is the probability that at least $1$ red ball is chosen?

$\displaystyle\frac{\binom72\cdot\binom41}{\binom{11}2}=\frac{28}{55}$

(b) What is the probability that the second ball chosen is red, given the that the first ball chosen was red?

$\displaystyle\frac{\binom72}{\binom{11}2}=\frac{21}{55}$

(c) What is the probability that the second ball chosen is red, given that the first ball chosen was not red?

$\displaystyle1-\frac{\binom71\cdot\binom41}{\binom{11}2}=\frac{27}{55}$

Answer 1

(a) The probability that no red balls are chosen, i.e. the probability that two blue balls are chosen, is $$\frac{4}{11}\frac{3}{10}$$ The probability that at least one red ball is chosen is therefore $$1-\frac{4}{11}\frac{3}{10}$$

(b) If the first ball was chosen red, there are then $6$ more red balls out of $10$ total, so the probability of selecting a red ball second is $\frac{6}{10}$

(c) Similar to (b). If the first ball chosen was blue, there are then $7$ red balls remaining and $10$ total, so the probability of selecting a red ball second is $\frac{7}{10}$

Answer 2

for the first part (A) you may ask: 2 balls are drawn from the urn (7 red, 4 blue) what is the probability that at least one is red (or in other words RR, OR ( RB/ BR )).

so you can say, either: RR

$\displaystyle\frac{\binom72\cdot\binom40}{\binom{11}2}$

OR (+) RB/BR

$\displaystyle\frac{\binom71\cdot\binom41}{\binom{11}2}$

and together:

$\displaystyle\frac{\binom72\cdot\binom40+\binom71\cdot\binom41}{\binom{11}2}$

for part b: you're left with 10 balls and you have 6 red - choose one.

now you can paraphrase C based on B.

Tip: this question falls under the hypergeometric probability category