Abox contains four $1 and six $5 bills. if three bills are selected at random without replacement, find the probability that all three are $5 bills.
a.27/125 b.1/4 c.1/6 d3/5

C.1/6

Step-by-step explanation:

Initially the box has four $1 and six $5 bills. The probability of selecting a $5 bill in the first trial would be given as;

(number of $5 bills) / (total number of bills)

= (6)/(4+6) = 3/5

If in the first attempt we actually pick a $5 bill, the number of $5 bills will reduce by one to 5. Now, the probability of picking a $5 bill in the second attempt will be given as;

(new number of $5 bills) / (new total number of bills)

= (5)/(4+5) = 5/9

The new number of $5 bills will now be; 6 - 2 = 4 since we have already picked 2 without replacing them.

Now, the probability of picking a $5 bill in the third attempt will be given as;

(new number of $5 bills) / (new total number of bills)

= (4)/(4+4) = 1/2

Since the three attempts are independent, the probability of picking  all three $5 bills is;

3/5 * 5/9 * 1/2 = 1/6

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.

all it means is function 3