Question

A bag contains 4 green, 3 black and 6 orange cards. A card sis drawn from the bag and kept, and then another card is drawn. Find the probability that
A, both cards will be orange
B, both cards will be black
C, the first card will be orange and the second will be black
D, neither will be orange

Answer 1

A. $P\left(both\:cards\:will\:be\:orange\right)=\frac{6}{13}\times \frac{5}{12}=\frac{5}{26}$

B. $P\left(both\:cards\:will\:be\:black\right)=\frac{3}{13}\times \frac{2}{12}=\frac{1}{26}$

C. $P\left(first\:card\:orange\:and\:second\:black\right)=\frac{6}{13}\times \frac{3}{12}=\frac{3}{26}$

D. The probability that neither of the cards will be orange means that either

(I). $P\left(both\:cards\:will\:be\:black\right)=\frac{3}{13}\times \frac{2}{12}=\frac{1}{26}$

or

(II). $P\left(both\:cards\:will\:be\:green\right)=\frac{4}{13}\times \frac{3}{12}=\frac{1}{13}$

or

(III). $P\left(first\:card\:will\:be\:green\:and\:second\:black\right)=\frac{4}{13}\times \frac{3}{12}=\frac{1}{13}$

or

(IV). $\:\:P\left(first\:card\:will\:be\:black\:and\:second\:green\right)=\frac{3}{13}\times \:\frac{4}{12}=\frac{1}{13}$

$P\left(neither\:of\:the\:cards\:will\:be\:orange\right)=\frac{1}{26}+\frac{1}{13}+\frac{1}{13}+\frac{1}{13}=\frac{7}{26}$