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Probability in Marbles | AMC 10A, 2010| Problem No 23

Try this beautiful Problem on Probability in Marbles based on smallest value AMC 10 A, 2010. You may use sequential hints to solve the problem.

Probability in Marbles – AMC-10A, 2010- Problem 23

Each of 2010 boxes in a line contains a single red marble, and for $1 \leq k \leq 2010$, the box in the $k$ th position also contains $k$ white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let $P(n)$ be the probability that Isabella stops after drawing exactly $n$ marbles. What is the smallest value of $n$ for which $P(n)<\frac{1}{2010}$ ?

,

• $20$
• $22$
• $44$
• $45$
• $46$

Probability

Combination

Marbles

Suggested Book | Source | Answer

Pre College Mathematics

Source of the problem

AMC-10A, 2010 Problem-23

Check the answer here, but try the problem first

$45$

Try with Hints

First Hint

Given that Each of 2010 boxes in a line contains a single red marble, and for $1 \leq k \leq 2010$, the box in the $k$ th position also contains $k$ white marbles..

Therefore The probability of drawing a white marble from box $k$ is $\frac{k}{k+1}$ and the probability of drawing a red marble from box $k$ is $\frac{1}{k+1}$

Now can you finish the problem?

Second Hint

Also given that She stops when she first draws a red marble. Let $P(n)$ be the probability that Isabella stops after drawing exactly $n$ marbles.

Therefore we can say $P(n)=\left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{n-1}{n}\right) \cdot \frac{1}{n+1}=\frac{1}{n(n+1)}$

Now Can you finish the Problem?

Third Hint

Therefore the probability $\frac{1}{n(n+1)}<\frac{1}{2010}$ or $n(n+1)>2010$

Now $n^2+n-2010>0$

Now to find out the factorization we see that $45 \times 46=2070$ and $44 \times 45 =1980$

As $n$ is smallest so $n=45$

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Probability in Game | AMC-10A, 2005 | Problem 18

Try this beautiful problem from AMC 10A, 2005 based on Probability in Game.

Probability in Game – AMC-10A, 2005- Problem 18

Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?

• $\frac{1}{4}$
• $\frac{1}{6}$
• $\frac{1}{5}$
• $\frac{2}{3}$
• $\frac{1}{3}$

Key Concepts

Probability

combinatorics

Answer: $\frac{1}{5}$

AMC-10A (2005) Problem 18

Pre College Mathematics

Try with Hints

Given that  The first team to win three games wins the series, team B wins the second game and team A wins the series. So the Total number of games played=$5$. Now we have to find out the possible order of wins…..

Can you now finish the problem ……….

Possible cases :

If team B won the first two games, team A would need to win the next three games. Therefore the possible order of wins is BBAAA.
If team A won the first game, and team B won the second game, the possible order of wins is $A B B A A, A B A B A,$ and $A B A A X,$ where $X$ denotes that the 5th game wasn’t played.
since ABAAX is dependent on the outcome of 4 games instead of 5, it is twice as likely to occur and can be treated as two possibilities.

According to the question, there is One possibility where team $\mathrm{B}$ wins the first game and 5 total possibilities, Therefore the required probability is $\frac{ 1}{5}$

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Probability Problem | AMC 8, 2016 | Problem no. 21

Try this beautiful problem from Probability.

Problem based on Probability | AMC-8, 2016 | Problem 21

A box contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?

• $\frac{3}{5}$
• $\frac{2}{5}$
• $\frac{1}{4}$

Key Concepts

probability

combination

fraction

Answer: $\frac{2}{5}$

AMC-8, 2016 problem 21

Challenges and Thrills in Pre College Mathematics

Try with Hints

There are 5 Chips, 3 red and 2 green

Can you now finish the problem ……….

We draw the chips boxes in such a way that we do not stop when the last chip of color is drawn. one at a time without replacement

Can you finish the problem……..

There are 5 Chips, 3 red and 2 green

we draw the chips boxes in such a way that we do not stop when the last chip of color is drawn.

if we draw all the green chip boxes then the last box be red or if we draw all red boxes then the last box be green

but we draw randomly. there are 3 red boxes and 2 green boxes

Therefore the probability that the 3 reds are drawn=$\frac{2}{5}$

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Problem from Probability | AMC 8, 2004 | Problem no. 21

Try this beautiful problem from Probability from AMC 8, 2004.

Problem from Probability | AMC-8, 2004 | Problem 21

Spinners A and B  are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners’ numbers is even?

• $\frac{2}{3}$
• $\frac{1}{3}$
• $\frac{1}{4}$

Key Concepts

probability

Equilly likely

Number counting

Answer: $\frac{2}{3}$

AMC-8, 2004 problem 21

Challenges and Thrills in Pre College Mathematics

Try with Hints

Even number comes from multiplying an even and even, even and odd, or odd and even

Can you now finish the problem ……….

A odd number only comes from multiplying an odd and odd…………..

can you finish the problem……..

We know that even number comes from multiplying an even and even, even and odd, or odd and even

and also a odd number only comes from multiplying an odd and odd,

There are few cases to find the probability of spinning two odd numbers from  1

Multiply the independent probabilities of each spinner getting an odd number together and subtract it from  1 we get…….

$1 – \frac{2}{4} \times \frac{2}{3}$= $1 – \frac{1}{3} = \frac{2}{3}$

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Probability | AMC 8, 2004 | Problem no. 22

Try this beautiful problem from Probability .You may use sequential hints to solve the problem.

Probability | AMC-8, 2004 |Problem 22

At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is$\frac{2}{5}$. What fraction of the people in the room are married men?

• $\frac{3}{8}$
• $\frac{1}{2}$
• $\frac{1}{4}$

Key Concepts

probability

combination

Number counting

Answer: $\frac{3}{8}$

AMC-8, 2007 problem 24

Challenges and Thrills in Pre College Mathematics

Try with Hints

Find the married men in the room …

Can you now finish the problem ……….

Find the total people

can you finish the problem……..

Assume that there are 10 women in the room, of which $10 \times \frac{2}{5}$=4 are single and 10-4=6 are married. Each married woman came with her husband,

so there are 6 married men in the room

Total man=10+6=16 people

Now The fraction of the people that are married men is $\frac{6}{16}=\frac{3}{8}$

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Probability | AMC 8, 2010 | Problem no. 24

Try this beautiful problem from Probability .You may use sequential hints to solve the problem.

Probability | AMC-8, 2007 |Problem 24

A bag contains four pieces of paper, each labeled with one of the digits 1,2,3 or 4.  with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of 3?

• $\frac{3}{4}$
• $\frac{1}{2}$
• $\frac{1}{4}$

Key Concepts

probability

combination

Number counting

Answer: $\frac{1}{2}$

AMC-8, 2007 problem 24

Challenges and Thrills in Pre College Mathematics

Try with Hints

there are two ways that The combination of digits that give multiples of 3

Can you now finish the problem ……….

The combination of digits that give multiples of 3 are (1,2,3) and (2,3,4)

can you finish the problem……..

The combination of digits that give multiples of 3 are (1,2,3) and (2,3,4) . The number of ways to choose three digits out of four is 4. Therefore, the probability is $\frac{1}{2}$