Suppose that distinct integer values are written on each of 3 cards. These cards are then randomly given the designations A, B, and C. The values on cards A and B are then compared.
1. If the smaller of these values is then compared with the value on card C, what is the probability that it is also smaller than the value on card C?

2/3

Step-by-step explanation:

The possible outcome from the three designations is express as 3!

and 3!=3x2x1=6

Hence the 6 possible outcome are listed below

(A,B,C),(A,C,B),(B,A,C),(B,C,A),(C,A,B) and (C,B,A)

let assume A should be less than B and B less than C, i.e A<B<C

The question required us to solve for

P(min(A,B)<C). And this can be expressed as

P(min(A,B)<C)=P(A<C)+P(B<C)-P(AB<C).

From the order given above, there are 3 possible outcome for A to come before C

Hence P(A<C)=3/6=1/2

also there 3 possible outcome for B to come first before C

P(B<C)=3/6=1/2

Also, there 2 events for A,B to come before C

P(AB<C)=2/6=1/3

Hence in final

P(min(A,B)<C)=P(A<C)+P(B<C)-P(AB<C)=1/2+1/2-1/3=2/3

the probability that it is also smaller than the value on card C is 2/3

what are the statements

step-by-step explanation:

step-by-step explanation:

i believe its "c" because all of the other options include a but it could be d because there is a way of financing with cash at certain places but its uncommon.