Exercise 21 Page 841

We want to find the probability that we hold a winning ticket. To do so, we will use theoretical probability. P=Favorable Outcomes/Possible Outcomes We will start by finding the number of possible outcomes. We know that lottery numbers are selected at random. We know the order the numbers are drawn in is not important, since the 6 numbers are drawn simultaneously. Therefore, the total number of possible outcomes will be the number of combinations of 6 of the 50 possible numbers. _nC_r = n!/r! (n-r)! In the formula for the number of combinations, n is the total number objects taken r at a time, where r ≤ n. In our case, the numbers are drawn from a pool of 50 possible numbers, so n = 50. Out of these, 6 numbers are drawn, so we know that r = 6. Let's substitute these values into our formula and find the number of possible combinations.

_nC_r = n!/r!(n-r)!

SubstituteII

n= 50, r= 6

_(50) C_6 = 50!/6!( 50- 6)!

▼

Evaluate right-hand side

SubTerm

Subtract term

_(50)C_6 = 50!/6! 44!

Write as a product

_(50)C_6 = 50 * 49 * 48 * 47 * 46 * 45 * 44 * 43 * ... * 2 * 1/(6 * 5 * 4 * 3 * 2 * 1)(44 * 43 * ... * 2 * 1)

CancelCommonFac

Cancel out common factors

_(50)C_6 = 50 * 49 * 48 * 47 * 46 * 45 * 44 * 43 * ... * 2 * 1/(6 * 5 * 4 * 3 * 2 * 1) (44 * 43 * ... * 2 * 1)

SimpQuot

Simplify quotient

_(50)C_6 = 50 * 49 * 48 * 47 * 46 * 45/6 * 5 * 4 * 3 * 2 * 1

Multiply

Multiply

_(50)C_6 = 11 441 304 000/720

CalcQuot

Calculate quotient

_(50)C_6 = 15 890 700

The total number of possible outcomes is 15 890 700. Next, we have to identify the number of favorable outcomes, where our one ticket matches the 6 drawn numbers exactly. There is only one combination of numbers that guarantees winning, so the number of favorable outcomes is 1. We are ready to calculate the desired probability. P=Favorable Outcomes/Possible Outcomes [0.9em] ⇕ P=1/15 890 700 The probability that we hold a winning ticket is 115 890 700.