Probability, Permutations, Combinations( jeff)

1. 1. A bag contains 6 marbles: 2 red marbles, 1 yellow marble, and 3 blue marbles. If you take one marble out, put it back, and take another marble out, what is the probability that you'll get 1 blue marble followed by 1 yellow marble 2.

1. A bag contains 6 marbles: 2 red marbles, 1 yellow marble, and 3 blue marbles. If you take one marble out, put it back, and take another marble out, what is the probability that you'll get 1 blue marble followed by 1 yellow marble? Solution Begin by noting that since the marbles are replaced, each draw does not depend on previous draws and thus the draws are independent. P(1 blue marble on first pick) = 3/6 = 1/2. P(1 yellow marble on second pick) = 1/6. P(picking 1 blue marble then picking 1 yellow marble) = (1/2) × (1/6) = 1/12

3. How many 7-digit telephone numbers can be formed using the digits 0 through 9? Notice that the first digit cannot be 0

3. How many 7-digit telephone numbers can be formed using the digits 0 through 9? Notice that the first digit cannot be 0 6 9*10. =9000000

I have anime lock with 3 dials that go 1-6 what are the total number of possibilities

6^3 = 216 Since there are 6 choices for the first number and six for the second and six for the third, you have 6 x 6 x 6 = 216 possible answers ranging from 111 112 113 . . . 664 665 666 So the "formula" is (number of choices) ^ (number of digits in the combination) If you have a lock with 3 digits and the possible choices 0-9 (which would be 10 numbers) then you would have 10^3 = 1000 possible combinations If the lock has 4 digits and the possible choices 0-9 (still ten numbers) the you would have 10^4 = 10000 possible combinations. See, math is useful! The area of mathematics that explains and uses counting arguments like these is called Combinatorics and is a really applicable subject. Helps count out possible card hands (like in poker) which helps you decide what is likely (probabilities). Sources: what's left of my brain unixcorn 100 months ago

Bill is making a sandwich. He may choose any combination of the following: lettuce, tomato, carrot, cheese, cucumber, beetroot, onion, ham. Find the probability that: a. the sandwich contains ham b. the sandwich contains three ingredients c. the sandwich contains at least three ingredients.

Bill is making a sandwich. He may choose any combination of the following: lettuce, tomato, carrot, cheese, cucumber, beetroot, onion, ham. Find the probability that: a. the sandwich contains ham b. the sandwich contains three ingredients c. the sandwich contains at least three ingredients Best Answer There are 8 ingredients. For each ingredient, Bill may either choose to have it, or to not have it. That means that there are 2^8 possibilities. That'll be the denominator in all of the sections a. How many ways can the sandwich contain ham? Well, it must contain ham, then of the other 7 options, bill can have it or not have it. That leads to 2^7/2^8 = 1/2 chance b. Bill can choose 3 ingredients in (8 choose 3) ways, or 8*7*6/(3*2*1)=8*7=56 ways. So 56/256=7/32 c. Bill can have 3 ingredients in (8c3), 4 ingredients in (8c4), etc. Another faster way to choose this is to just take 8c0+8c1+8c2 and subtract that from 2^8. That leads to 2^8-1-8-28 (do you see why we can do this?). 256-37=219 219/256 Elite · 5

How many outfits combinations does a Jimmy have if he has 4 shirts, 6 ties, and 7 pairs of pants.

How many outfits combinations does Jimmy have if he has 4 shirts, 6 ties, and 7 pairs of pants. Think of it this way: He has 4 ways to select a shirt and for each of those 4 ways to select a shirt, he has 6 ways to select a tie, giving you 24 shirt and tie combinations. Then, for each of those 24 shirt and tie combinations, you have 7 pairs of pants to choose! 168 possibilities.

In how many ways can the letters in MISSISSIPPI be arranged? Suppose the 2 P's must be separated?

In how many ways can the letters in MISSISSIPPI be arranged? Suppose the 2 P's must be separated? Including a place at the beginning and end of the row we must select two positions to put the p's from the 10 positions available. These two positions can be chosen in C(10,2) = 45 ways. The other 9 letters could be arranged in 9!/(4! 4!) = 630 ways. So the total number of ways is 45 x 630 = 28350 ways. An alternative method is to stick the two p's together and find the number of arrangements when they are together. We have 10 objects to permute and the number of different arrangements is 10!/(4! 4!) = 6300 Subtract this from the total of all possible arrangements, 34650, to get the number of arrangements with the two p's separated 34650 - 6300 = 28350

Mark has 5 pants and 7 shirts in his closet. He wants to wear a different pant/shirt combination each day without buying new clothes for as long as he can. How many weeks can he do this for?

Mark has 5 pants and 7 shirts in his closet. He wants to wear a different pant/shirt combination each day without buying new clothes for as long as he can. How many weeks can he do this for? The fundamental counting principle says that if you want to determine the number of ways that two independent events can happen, multiply the number of ways each event can happen together. In this case, there are 5 * 7, or 35 unique combinations of pants & shirts Mark can wear. If he wears one combination each day, he can last 35 days, or 5 weeks,

Two 6-sided dice are rolled. Find the odds of rolling s sum that is a prime number

Two 6-sided dice are rolled. Find the odds of rolling s sum that is a prime number Possible prime sums are 2,3,5,7, and 11 number of ways to get 2 = 1 number of ways to get 3 = 2 number of ways to get 5 = 4 number of ways to get 7 = 6 number of ways to get 11 = 2 total = 15 prob of that happening = 15/36 = 5/12 so prob of that NOT happening = 1-5/12 = 7/12 odds in favor of a sum which is prime = (5/12) : (7/12) = 5

Two dice are rolled. Find the probability that an even number is rolled on one die and an odd number is rolled on the second die

Two dice are rolled. Find the probability that an even number is rolled on one die and an odd number is rolled on the second die make a chart with DIE 1 on top and DIE 2 on the side. there will be 36 outcomes, 18 even sums and 18 odd sums. (notice that the sum of an even + odd = odd, but the sum of either two evens or two odds is even) so prob = 18/36 = 1/2

What is the probability of rolling a 6 with a pair of dice?

What is the probability of rolling a 6 with a pair of dice? There are five ways to rolls a six: 1-5,2-2,3-3,4-2,5-1. There are 6 squared=36 outcomes when a pair of dice is rolled. The probablity of rolling a 6 is 5/36

What is the probability of selecting a heart from a shuffled standard deck of cards?

What is the probability of selecting a heart from a shuffled standard deck of cards? There are 13 hearts in a standard deck of 52 cards so the probability is 13/52 or 1/4

What is the probability that a randomly selected permutation of the letters G, G, I, W, L, E would spell "wiggle"? A. 1/720 B. 1/360 C. 1/180 D. 1/90

What is the probability that a randomly selected permutation of the letters G, G, I, W, L, E would spell "wiggle"? A. 1/720 B. 1/360 C. 1/180 D. 1/90 You have 6 letters (let's call them G1, G2, I, W, L, and E so we can tell the G's apart). Thus, there are 6!=6×5×4×3×2×1=720 6 6 5 4 3 2 1 720 permutations of them. Of those 720 permutations, precisely 2 of them spell the word "wiggle": WIG1G2LE and WIG2G1LE. Hence, your probability is 2/720=1/360 2 720 1 360 .

What is the probability that the position in which the consonants appear remain unchanged when the letters of the word "Math" are re-arranged?

What is the probability that the position in which the consonants appear remain unchanged when the letters of the word "Math" are re-arranged? The total number of ways in which the word Math can be re-arranged = 4! = 4*3*2*1 = 24 ways. Now, if the positions in which the consonants appear do not change, the first, third and the fourth positions are reserved for consonants and the vowel A remains at the second position. The consonants M, T and H can be re-arranged in the first, third and fourth positions in 3! = 6 ways without the positions in which the positions in which the consonants appear changing. Therefore, the required probability =1/4

What's is the probability that the two top cards are face cards?

What's is the probability that the two top cards are face cards? There are 3 face cards in each suit for a total of twelve face cards in a deck of 52 the probability the lag the top card is a face card is 12/52 and if the top card is a face card the second face card will have the probability of 11/51 the probability of both cards being face cards is 12/52*11/51=11/221

candy that had 10 pieces of Milky Way, Almond Joy, Butterfinger, Nestle Crunch, Snickers, and Kit Kat. Jen grabbed six pieces at random from this bowl of 60 chocolate candies. (a) What is the probability that she got one of each variety?

candy that had 10 pieces of Milky Way, Almond Joy, Butterfinger, Nestle Crunch, Snickers, and Kit Kat. Jen grabbed six pieces at random from this bowl of 60 chocolate candies. (a) What is the probability that she got one of each variety? each type of candy, there's 10 items, and you choose 1. You do this six times.

notebook Example 2: In how many ways can 8 CD's be arranged on a shelf?

notebook Example 2: In how many ways can 8 CD's be arranged on a Shelf Expand out 8 = 40320