When Will Petr Win?

When Petr plays Game 1, 2, 3, 4, 5, 6, 7, 8, 9 or 10 against Diego, he has a $\frac {1}{100}$ , $\frac {4}{100}$ , $\frac {9}{100}$ , $\frac {16}{100}$ , $\frac {25}{100}$ , $\frac {36}{100}$ , $\frac {49}{100}$ , $\frac {64}{100}$ , $\frac {81}{100}$ , or $\frac {100}{100}$ chance of winning respectively. However, a fair 10-sided dice is rolled to determine the game that is played. Thus the chance of playing and winning a game is $\frac {i^2}{100 \times 10}$ .

By the Rule of Sum, the probability of Petr winning against Diego is thus $\frac {1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2}{100 \times 10} = \frac {77}{200}$

Thus, $a+b=277$

The probability that he wins is the summation of (probability of occurrence of ith game x probability of winning that game). Therefore the probability is: 0.1( $1^2$ /100)+0.1( $2^2$ /100)+0.1( $3^2$ /100)+......+0.1( $10^2$ /100) which is equal to 0.1 ((10)(10+1)(2(10)+1)/6)/100 (using sum of squares of first 10 natural numbers) which simplifies down to 385/1000 or 77/200.

the probability that game $i$ will be selected is $\tfrac{1}{10}$ for $i=1, 2, ..., 10$ . Hence, the probability that Peter will win against Diego if $\tfrac{1}{10}\left(\dfrac{1^2}{100} + \dfrac{2^2}{100} + \cdots + \dfrac{10^2}{100}\right) = \dfrac{1}{10}\cdot \dfrac{10(11)(21)}{6\cdot100} = \dfrac{385}{1000} = \dfrac{77}{200}$ . Thus, $a=77$ and $b=200$ . Therefore, $a+b=277$

You must add the probability of winning each game. For example, in game 3, i =3, so the probability of winning is (i^2)/100, or 9/100. So (1/100) + (4/100) + (9/100) + (16/100) + (25/100) + (36/100) + (49/100) + (64/100) + (81/100) + (100/100) = (385/100). Since there are ten different games, each with equal probability of being chosen, to find the average you would have to divide this number by 10. (385/100)/10 is equal to (385/1000), which is simplified to (77/200), which is written in the form of a/b. Therefore, a+b would equal 200 + 77, or 277.

By the addition rule, we have probability of winning of $10$ games = $\frac{\sum_{i=1}^{10} \frac{i^2}{100}}{10}$

$= \frac{\frac{1^2}{100} + \frac{2^2}{100} + \frac{3^2}{100} + \dots + \frac{10^2}{100}}{10}$ $= \frac{1^2 + 2^2 + 3^2+ \dots + 10^2}{1000}$

The sum of squares of the first $n$ natural numbers is given by $\frac{ n(n+1)(2n+1)}{6}$

$\therefore P(\text{winning}) = \frac{10 \times 11 \times 21}{6 \times 1000} = \frac{77}{200}$

So the answer is $77 + 200$ which is equal to $277$ .

There is a 1/10 chance of playing each game, and i^2/100 chance of winning each game. Working out sum(i^2) for i=1 to 10 comes to 385, and this needs to be divided by 1/10th of 1/100, or 1/1000. This makes 385/1000, which reduces to 77/200. The answer is 277.

We first count the probability of playing a certain game. Note that No matter which game is to be played, the probability is $$\frac{1}{10}$$. Now that we have selected a game to be played, We can continue computing the probability of Petr to win that game which is given as $$\frac{i^2}{100}$$ for $$ i=1,2,3,....10$$. So we add all such probabilities to get the desired answer which is $$ \frac{1}{10}\frac{1}{100} + \frac{1}{10}\frac{2^2}{100} + ..... +\frac{1}{10}\frac{10^2}{100}\Longleftrightarrow\frac{1}{10}\frac{385}{100}=\frac{77}{200}=\frac{a}{b}$$ Using the well known identity, $$ 1^2 + 2^2 +........+n^2 = \frac{n(n+1)(2n+1)}{6}$$ so, $$a+b=277$$

Petr's probability of winning is the sum of the probability that he wins each game multiplied by the probability that he and Diego play that game. Since a fair die is rolled, the probability that he plays each game is $\frac{1}{10}$ . Thus, the probability Petr wins is $\sum_{i = 1}^{10}\frac{i^2}{1000}$ . We can use the sum of squares formula to get that this is $\frac{(10)(11)(21)}{(6)(1000)} = \frac {11 \times 7} { 2 \times 100} = \frac{77}{200}$ . Therefore, $a + b = 77 + 200 = 277$ .

There is a \frac {1}{10} or 10% chance that Petr picks each individual game. In addition, the COMBINED probability of winning all ten games would be \frac {1 + 4 + 9.. + 81 + 100}{100} . Multiplying that by .1 (the chance he picks each game), you get a final fraction of \frac {77}{200} . 200 + 77 equals 277.