A prize machine randomly dispenses small toys. There are 1515 prizes left:
55 bouncy balls, 33 toy cars, 22 stickers, and 55 dinosaurs.
Kai gets 22 prizes. What is the probability that they are both toy cars?
1/35
There are 3C2 ways in getting both toy cars, and our sample space is 15C2.
Thus, the probability becomes: 3C2/15C2 = 1/35
A prize machine randomly dispenses small toys. There are 15 prizes left:
5 bouncy balls, 3 toy cars, 2 stickers, and 5 dinosaurs.
Kai gets 2 prizes. What is the probability that they are both toy cars?
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
3/15 * (3-1)/(15-1) = ?
possibilities are
A) 3/70
B) 3/75
C) 1/35
D) 2/70
PsyDAG is right
To find the probability that both prizes are toy cars, we need to first determine the total number of possible outcomes, which is the total number of prizes left (1515).
Next, we need to determine the number of favorable outcomes, which is the number of ways we can select 2 toy cars from the remaining prizes. Since there are 33 toy cars remaining, we can use the formula for combinations to calculate this. The formula for combinations is nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items we are selecting. In this case, n = 33 and r = 2.
Using the formula, the number of ways to select 2 toy cars from 33 is:
33! / (2!(33-2)!) = 33! / (2! * 31!) = (33 * 32) / 2 = 528.
Therefore, the probability that both prizes are toy cars is:
Number of favorable outcomes / Total number of possible outcomes = 528 / 1515.
To calculate this probability, divide 528 by 1515 using a calculator or perform long division.