Question

What is the probability of choosing a pair that contains ONLY cards labeled with the letters B, C, or F?

Answer 1

Answer:

$Pr = \frac{1}{6}$

Step-by-step explanation:

Given

$Cards = \{A,B,C,D,E,F\}$

$r = 2$ --- selection

Required

Probability of choosing a pair with only B, C and F

The total selection is: 36

This is calculated as thus:

$First = 6$ --- First selection can be any of the 6 cards

$Second = 6$ --- Second selection can be any of the 6 cards [No restriction]

So, we have:

$n = First * Second$

$n = 6 * 6$

$n = 36$

The pair of only B, C or F is:

$Pair = \{(B,C), (B,F), (C,B),(C,F),(F,B),(F,C)\}$

$n(Pair) = 6$

So, the probability is:

$Pr = \frac{n(Pair)}{n}$

$Pr = \frac{6}{36}$

Simplify

$Pr = \frac{1}{6}$