Question

(a) Two numbers are chosen at random from among the numbers 1 to 10 without

replacement. Find the probability that the second number chosen is 5.

(b) A box contains 8 yellow, 5 green and 7 black balls identical to each other except for colour. 3 balls are drawn at random one after the other without replacement. Find the probability that:

(i) there will be one ball of each colour.

(ii)exactly one ball will be black

(iii) at least one ball will be black

(iv) the second ball will be black

(v) the second ball will be black given that the first is black.

Answer 1

(a). The probability that the second number will be chosen is 5$=\frac{\left(9C_1\times 1\right)}{\left(10C_2\times 2!\right)}=0.1$

(b).

(i). The probability that there will be one ball of each color $=\frac{\left(8C_1\times 5C_1\times 7C_1\times 3!\right)}{\left(20C_3\times 3!\right)}=0.2456$

(ii). The probability that there will exactly one ball will be black $=\frac{\left(7C_1\times 13C_2\right)}{20C_3}=0.4789$

(iii). The probability that there will at least one ball will be black $=1-P\left(no\:black\right)$

$=1-\frac{13C_3}{20C_3}=0.2509$

(iv). The probability that the second ball will be black $=\frac{\left(7C_1\times 19C_2\times 2!\right)}{\left(20C_3\times 3!\right)}=0.35$

(v). The probability that the second ball will be black given that the first one is black $=\frac{6}{19}=0.3158$