How do I solve these problems and what are the answers?

Question

In a lottery, you must choose three different numbers (1-30). To win, your numbers must match the three numbers drawn in any order. Numbers cannot be repeated. How many different lottery tickets are possible?

Mr. Starnes puts the names of all 15 of his 4th period students on popsicle sticks and places them in a jar. When he wants to call on someone, he chooses a popsicle stick. He always replaces the popsicle stick before he draws another name. If Mr. Starnes chooses two students while checking homework, what is the probability that Ashlyn will be called on at least once?

If you flip a coin five times, how many different outcomes are possible?

 

When flipping a coin three times, what is the probability that you will get at least one tail?

When rolling two six-sided dice, find P(even sum or sum less than 5).

There are eight people running to be a student council representative. Three representatives will be chosen. How many different ways can the three representatives be chosen?

 

A drug dealer has hidden drugs in barrels of chemicals being shipped to the Czech Republic. The shipping container used holds 50 barrels. Only six of the barrels actually contain drugs. When checking the container, a customs agent randomly selects three barrels to check for drugs (without replacement). What is the probability that he will choose at least one barrel with drugs in it?



Answer 1

(a) In drawing the winning lottery numbers, there will be 30 possible choices for the first number. For each of those 30 possibilities, there will be 29 possible choices for the second number, giving a total of 30*29 = 870 possible two-number sequences. Finally, for each of those 870 two-number sequences, there are 28 possible choices for the third number. This means that there are a total of 870*28 = 24,360 possible three-number sequences.

However, the question tells us that the order of the numbers doesn't matter. For instance, (1, 2, 3) and (3, 2, 1) give the same lottery numbers, but are counted separately in the 24,360 figure. To account for this, we must divide 24,360 by the number of possible ways to arrange three numbers. If we have 3 numbers and we want to arrange them in some order, then the first number in the sequence can be any of those 3 numbers. After choosing the first number, the second number can be any of the 2 remaining numbers. Finally, the last number is the only remaining number. This gives a total of 3*2*1 = 6 orderings.

Putting this together, there are 24,360/6 = 4,060 possible lottery tickets.

(b) This question can be more easily solved by finding the probability that Ashlyn will not be called on at least once. Then, since there is a 100% chance that she is either called on or not, we can subtract this value from 1 to get the desired probability.

In the first selection, 14 out of the 15 popsicle sticks do not have Ashlyn's name written on them. So, there is a 14/15 chance that Mr. Starnes will not choose Ashlyn the first time. However, Mr. Starnes puts the popsicle sticks back into the cup! So the second time, there will again be a 14/15 chance that Mr. Starnes will not choose Ashlyn. Together, there is a (14/15) * (14/15) = 196/225 ≈ 0.871 = 87.1% chance that Mr. Starnes doesn't choose Ashlyn in either selection.

Thus, the probability that Ashlyn is called in at least one of these selections is 1 - 196/225 = 29/225 ≈ 0.128 = 12.8%

(c) Each coin flip, there are two possible outcomes: heads or tails. So, for the first flip, there are two possibilities. For each of those possibilities, there are two possibilities for the second flip, giving a total of 2*2 = 4 possible outcomes for two flips. Repeating this process, the total number of possible sequences of five coin flips is 2⁵ = 32.

(d) We will use the same strategy as in (b): find the probability of getting no tails, then subtracting this from 1. Using the idea from (c), we find that the total number of possible sequences for three coin flips is 2³ = 8. Only one of these has no tails. Namely, the sequence HHH (consisting entirely of heads). Thus, the probability of flipping no tails is 1/8 = 0.125 = 12.5%. Therefore, the probability of getting at least one tail is 1 - 1/8 = 7/8 = 0.875 = 87.5%

(e) We can compute this probability by splitting it into parts: P(sum is even or less than 5) = P(sum is even) + P(sum is less than 5) - P(sum is even and less than 5). That last term is necessary because the sum could be 2, for instance. This possibility would be double-counted without subtracting away that last term.

First, we find the probability that sum is even. Since a die has six sides, three of which are even and three of which are odd, a given dice roll will be even 50% of the time and odd 50% of the time. To get an even sum, both dice must be the same parity (i.e., both dice must be even or both dice must be odd). There are two other possibilities, however: the first dice could be even and the second odd, or the first could be odd and the second even. All four of these possibilities are equally likely, and two of them yield an even sum. Hence we have a 50% = 1/2 chance that the sum will be even.

Second, we find the probability that the sum is less than 5. The possible rolls which could yield such a sum are (1,1), (1,2), (2,1), (1,3), (2,2), and (3,1), giving us six total rolls which have a sum less than 5. Each die has six possibilities, so there are 6*6 = 36 total rolls. Hence, the probability of rolling a sum less than 5 is 6/36 = 1/6.

Finally, the probability that sum is less than 5 and even is found by taking the probability that the sum is less than 5 and multiplying it by the probability that the sum is even given that the sum is less than 5. We already know that there are 6 possible ways to roll a sum less than 5. Of those 6 possibilities, 4 of them have an even sum: (1,1), (1,3), (2,2), (3,1). Hence, the probability that the sum is even given that the sum is less than 5 is 4/6 = 2/3. Putting these together, the probability that the sum is less than 5 and even is (1/6)*(2/3) = 2/18 = 1/9.

Putting everything together, P(sum is even or sum is less than 5) = 1/2 + 1/6 - 1/9 = 5/9 ≈ 55.6%

(f) We'll assume that the representatives have the same title so that order doesn't matter. Then we can follow the exact same procedure as in (a). There are 8 possible candidates to choose for the first representative, then 7 possible choices for the second representative, and 6 possible choices for the third representative, for a total of 336 possible ways to choose three representatives in order. However, we assume that order doesn't matter, so we must divide by the number of ways to order these three representatives. In part (a), we computed that the number of ways to order three objects is 6. Hence, there are 336/6 = 56 ways to choose three representatives from 8 candidates.

(g) We use the same strategy as in (b) and (d): find the probability that they don't find any drugs, then subtract that value from 1.

For the first barrel, there are 44 possible choices that do not contain any drugs, meaning the customs agent has a 44/50 chance of selecting a barrel empty of drugs for the first choice. If any of these barrels are chosen, there are then 49 barrels to choose from, and 43 which do not contain drugs, giving a 43/49 chance of selecting a second barrel containing no drugs. Finally, if the customs agent again chooses a barrel without drugs, there are 48 barrels to choose from and 42 that do not contain drugs, giving a 42/48 chance that the customs agent doesn't choose a barrel with drugs for the third barrel.

The probability that these three possibilities occur consecutively is (44/50)*(43/49)*(42/48) = 473/700 ≈ 67.57%. Hence, the probability that the customs agent finds at least one barrel containing drugs is 1 - 473/700 = 227/700 ≈ 32.43%. Looks like it's in the drug dealers' favor!