 # High school probability questions and answers

Recent questions in High school probability syaoronsangelhwc17 2022-05-13 Answered

### A committee of 4 people is chosen from 8 women and 8 men. How many different committees are possible that consist of 2 women and 2 men? joyyy joy 2022-05-13

### Let A and B be events from a sample space S, with P[A] = 0.85 and P[B] = 0.8. Specify the range of possible values for P[A ∪ B] (for example: 0.3 ≤ P[A ∪ B] ≤ 0.9). dumnealorjavgj 2022-05-10 Answered

### A catering service offers 12 appetizers, 9 main courses, and 6 desserts. A customer is to select 5 appetizers, 3 main courses, and 4 desserts for a banquet. In how many ways can this be done? Reese Estes 2022-05-09 Answered

### How many ways can you buy 2 DVDs from a display of 15? Derick Richard 2022-05-09 Answered

### 20 blue balls and 11 yellow, drawing 6 times with no replacement, what is the chance that at least one is yellow OR the first two draws are the same?I'm not sure my intuition for the solution of the problem is correct, in particular, I have some issue with the addition rule part.I started out this problem by thinking about the two scenarios:1) the first two are yellow2) the first two are blueI'm a bit confused about the first scenario: P(at least 1 yellow) OR P(first two are the same) if the first two are yellow, that also satisfy the P(at least 1 yellow) so in this case does the probability equal to 1?Is this similar to flipping a coin P(head) or P(tail)? Or perhaps it's only a "double counted intersection" that we have to subtract out?I also thought about using compliment to solve this problem but I'm not quite sure if it works this way:1- [P(no yellow) or P(first two are different)] hisyhauttaq84w 2022-05-09 Answered

### What is $P\left(A\cup \left(B\cap C\right)\right)$?The question says it all. I know$P\left(A\cap \left(B\cup C\right)\right)=P\left(A\cap B\right)+P\left(A\cap C\right).$Would this mean,$P\left(A\cup \left(B\cap C\right)\right)=P\left(A\cup B\right)+P\left(A\cup C\right)?$Just want to make sure. Jaeden Weaver 2022-05-09 Answered

### How do you evaluate 10P8 using a calculator? dresu9dnjn 2022-05-09 Answered

### Out of a group of 21 persons, 9 eat vegetables, 10 eat fish and 7 eat eggs. 5 persons eat all three. How many persons eat at least two out of the three dishes?My approach: $N\left(A\cup B\cup C\right)=N\left(A\right)+N\left(B\right)+N\left(C\right)-N\left(A\cap B\right)-N\left(A\cap C\right)-N\left(B\cap C\right)+N\left(A\cap B\cap C\right)$$21=9+10+7-N\left(A\cap B\right)-N\left(A\cap C\right)-N\left(B\cap C\right)+5$$N\left(A\cap B\right)+N\left(A\cap C\right)+N\left(B\cap C\right)=10$Now the LHS has counted $N\left(A\cap B\cap C\right)$ three times, so I will remove it two times as:-Number of persons eating at least two dishes $=N\left(A\cap B+B\cap C+A\cap C\right)-2\ast N\left(A\cap B\cap C\right)=10-2\ast 5=0$Now it contradicts the questions that there are 5 eating all three dishes.Is this anything wrong in my approach? kazue72949lard 2022-05-09 Answered

### Alfonso and Colin each bought one raffle ticket at the state fair. If 50 tickets were randomly sold, what is the probability that Alfonso got ticket 14 and Colin got ticket 23?The answer should be $\frac{1}{2450}$ which presumably comes from $\frac{1}{50}×\frac{1}{49}$. But it seems that the order does not count. I did not assume that Alfonso got ticket 14 first then Colin got ticket 23 second.Update: What is wrong with this reasoning. When I said that I did not assume order, I meant that it's possible1. Alfonso got ticket 14 first, then Colin got ticket 23,2. Colin got ticket 23 first, then Alfonso got ticket 14.Both of these possibilities are possible before the tickets are given out, so we can make an 'or' statement. Label the event Alfonso got ticket 14 by ${A}_{14}$ and Colin got ticket 23 by ${A}_{23}$. Then by the addition ruleI realize that once the tickets are sold, then only one of must occur, but before the tickets are sold both possibilities are plausible. Why would the probability change before and after the tickets are sold. sembuang711q6 2022-05-09 Answered

### Getting 2 or 5 in two throws should be $P\left(2\right)+P\left(5\right)$. $P\left(2\right)=1/6,P\left(5\right)=1/6$ so the combined so it should be 1/3.I tried to visualize but not able to do so correctly.11,12,13,14,15,16, 21,22,23,24,25,26,31,32, ....6,6total of 36 possibilities.12,15,21,22,23,24,25,26,31,35,42,45,51,52,53,54,55,56,61,65out of which 20 possibilities, so the probability should be 20/36 which is not 1/3.Where am I going wrong? Jordon Haley 2022-05-09 Answered

### Let $A$ and $B$ be subsets of the finite set $S$ with $S=A\cup B$ and $A\cap B=\mathrm{\varnothing }$. Denote by $\mathcal{P}\left(X\right)$ the power set of $X$ and denote by $|Y|$ the number of elements in the set $Y$.Given a statement $|\mathcal{P}\left(A\right)|+|\mathcal{P}\left(B\right)|=|\mathcal{P}\left(A\right)\cup \mathcal{P}\left(B\right)|$Use the Addition Counting Principle to prove or disprove the statement.I understand that its asking me to find the elements of $\mathcal{P}\left(A\right)$ and $\mathcal{P}\left(B\right)$, but where does $\mathcal{P}\left(X\right)$ and $|Y|$ fit in to solve this question? Brody Collins 2022-05-09 Answered

### Let n be a positive integer. Find the number of permutations of $\left(1,2,...,n\right)$ such that no number remains in its original place.Solution: To do this, first we are going to count the number of permutations where at least one number remains in its place, according to the inclusion-exclusion principle, we must first add the permutations where a given number is fixed, then subtract the permutations where 2 given numbers are fixed and so on.To find a permutation that fixes $k$ given elements, we only have to arrange the rest, which can be done in $\left(n-k\right)!$ ways. However, if we do this for every choice of $k$ elements, we are counting $\left(\genfrac{}{}{0}{}{n}{k}\right)\left(n-k\right)!=\frac{n!}{k!}$ permutations. Since in total there are $n!$ permutations we get as our result:$n!-\left(\frac{n!}{1!}+\frac{n!}{2!}-\frac{n!}{3!}+...+\left(-1{\right)}^{n+1}\frac{n!}{n!}\right)$I'm a bit confused about a point of the solution, when we fix $k$ elements and rearrange the other $n-k$, some of the remaining elements will stay fixed in their position right? so why does this work? garcialdaria2zky1 2022-05-09 Answered

### a card is chosen from a standard deck of 52 cards, what is the probability that a spade and a heart is chosen 3c4ar1bzki1u 2022-05-08 Answered

### How would you use permutations to find the arrangements possible if a line has 4 girls and 4 boys alternating boy girl boy girl? vilitatelp014 2022-05-08 Answered

### How many different permutations are there of the letters in the word MISSISSIPPI? Jazlyn Raymond 2022-05-08 Answered

### 1. Sailing ships used to send messages with signal flags flown from their masts. How many different signals are possible with a set of four distinct flags if a minimum of two flags is used for each signals?2. A Gr. 9 students may build a timetable by selecting one course for each period, with no duplication of courses. Period 1 must be science, geography, or physical education. Period 2 must be art, music, French, os business. Period 3 and 4 must be math or English. How many different timetables could a student choose? Matilda Webb 2022-05-08 Answered

### Coach Ringdahl has 10 players on his varsity basketball team. How many different starting line-ups (5 players) can he choose? Derick Richard 2022-05-07 Answered

### i have a homework with 10 question but im stuck with 3 i searched about them everywhere read other colleges lectures but i couldnt solved them finally i desired to ask hereQuestion-2:Prove that each positive integer can be written in form of ${2}^{k}\ast q$, where q is odd, and k is a non-negative integer.Hint: Use induction, and the fact that the product of two odd numbers is odd.Question-6:$\left(x+y{\right)}^{n}=\sum _{k=0}^{n}C\left(n,k\right)\ast {x}^{n-k}\ast {y}^{k}=\frac{{n}^{2}+n}{2}$Prove the above statement by using induction on n.Question-7:Let ${n}_{1},{n}_{2},...,{n}_{t}$ be positive integers. Show that if ${n}_{1}+{n}_{2}+...+{n}_{t}-t+1$ objects are placed into t boxes, then for some i, $i=1,2,3,...,t$ , the ith box contains at least ${n}_{i}$ objects. mars6svhym 2022-05-07 Answered

### How many 5-digit numbers can be formed using $\left(0-9\right)$?

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