Video Transcript
Elizabeth designed the following simulation to model the outcomes of a game at a fair. Each trial will model one round of the game. A trial will consist in randomly generating a number between one and 10. The number one will represent winning first prize, the numbers two and three second prize, the numbers four, five, and six, third prize, and the rest of the numbers will represent losing the game. What is the theoretical probability of winning the second prize in a game?
There is also a second and third part to this question that we will look at later.
The theoretical probability of an event occurring can be written as a fraction, the number of successful outcomes over the number of possible outcomes. When playing the game, a person could win first prize, second prize, third prize, or lose the game. As each trial will randomly generate a number between one and 10, there are 10 possible outcomes.
A person wins first prize if they land on the number one. This means that the probability of winning first prize is one out of 10 or one-tenth. Landing on a two or three means that the person would win second prize. The probability of winning second prize is therefore two out of 10 or two-tenths. Landing on four, five, or six win third prize. As there are three successful outcomes here, the probability of third prize is three-tenths. The numbers seven, eight, nine, and 10 are remaining. As there are four numbers here, the probability of losing the game is four-tenths.
In this question, we are asked to work out the probability of winning second prize. Two-tenths is equivalent to 0.2 as two divided by 10 is 0.2. As a percentage, this is equal to 20 percent. To convert a decimal to a percentage, we multiply by 100. We could also have simplified the fraction to one-fifth by dividing the numerator and denominator by two. Any of these four answers would be correct. The theoretical probability as a percentage of winning second prize is 20 percent.
We will now look at the second part of this question.
She used a random number generator to simulate 50 games, and the frequencies of each number are given in the table. Find the experimental probability of winning the second prize.
We remember from the first part of the question that landing on the number one won first prize. Landing on two or three won second prize. Landing on the numbers four, five, or six won third prize. And finally, landing on seven, eight, nine, or 10 lost the game.
We can therefore split our table into four sections. As there were 50 games in total, the experimental probability of winning first prize is four out of 50. The probability of winning second prize was 17 over 50 as eight plus nine is equal to 17. The probability of winning third prize was nine out of 50 as three plus two plus four is equal to 9. And finally, the probability of losing the game was 20 out of 50. Five plus five is equal to 10, adding four gives us 14, and adding six gives us 20.
Once again, we’re interested in second prize. 17 out of 50 or seventeen fiftieths is the same as 34 out of 100. Remember, whatever we do to the numerator of our fraction we must do to the denominator. In this case, we’ve multiplied both by two. This is equivalent to 0.34 or 34 percent. The experimental probability of winning second prize is 34 percent.
We will now clear some space to answer the third part of the question.
Which bar graph summarizes the experimental probabilities that can be calculated as a result of her simulation?
We worked out in the last part of the question that the probability of winning second prize was 34 percent. This was because 17 out of 50 is equal to 34 percent. In the same way, four out of 50 is equal to 8 percent, nine out of 50 is equal to 18 percent, and 20 out of 50 is equal to 40 percent. Writing these percentages as decimals gives us 0.08, 0.34, 0.18, and 0.4 or 0.40. The only bar graph that shows a probability of first prize of 0.08 is bar graph A.
Likewise, bar graph A is the only one that shows the correct probability for second prize. The bar for third prize is correct in bar graph A, bar graph B, and bar graph D. This is also true for the probability of losing the game. The only bar graph that is correct for all four outcomes is bar graph A. Therefore, this is the one that summarizes the experimental probabilities.
