A bag contains the following prize tickets:%0D%0A%0D%0A • 6 tickets labeled “8-ounce bottle of SLIME”%0D%0A%0D%0A • 2 tickets labeled “4 RINGTONES” %0D%0A%0D%0A • 8 tickets labeled “32-ounce SLUSHIE” %0D%0A%0D%0A • 4 tickets labeled “16-ounce ICE CREAM SUNDAE”.%0D%0A %0D%0A%0D%0A%0D%0A%0D%0AEach of the 260 7th grade students who passed the Math benchmark test will draw a prize ticket from the bag. The prize will be recorded for that student, then the ticket will be returned to the bag. %0D%0A%0D%0AWhat is a reasonable prediction for the number of times a prize ticket for slime will NOT be drawn?
To find a reasonable prediction for the number of times a prize ticket for slime will NOT be drawn, we first need to calculate the total number of slime tickets in the bag:
6 tickets labeled “8-ounce bottle of SLIME”
Next, we calculate the total number of prize draws:
260 students who passed the Math benchmark test
Since each student will draw a prize ticket and the ticket will be returned to the bag after recording the prize, the total number of prize draws is equal to the total number of students, which is 260.
Now, we can find the probability of not drawing a slime ticket in one draw:
Probability of not drawing a slime ticket = 1 - Probability of drawing a slime ticket
Probability of drawing a slime ticket = Number of slime tickets / Total number of tickets
Probability of drawing a slime ticket = 6 / (6 + 2 + 8 + 4) = 6 / 20 = 0.3
Probability of not drawing a slime ticket = 1 - 0.3 = 0.7
Now, we can predict the number of times a prize ticket for slime will NOT be drawn:
Number of times a prize ticket for slime will NOT be drawn = Total number of prize draws x Probability of not drawing a slime ticket
Number of times a prize ticket for slime will NOT be drawn = 260 x 0.7 = 182
Therefore, a reasonable prediction is that a prize ticket for slime will NOT be drawn approximately 182 times.