Celebration is being held honoring five award winners of different ages each winner will be called one at a time in random order
1 answer:
Answer:
1 in 120 or 0.83%
Step-by-step explanation:
There are 5 possible winners to be called first, 4 possible winners are then left to be called second, 3 to be called third, 2 to be called fourth and the final winner is left to be called last. Therefore, the number of possible ways to arrange the winners is:
There is only one possible order for which the winners are called in order by the age from youngest to oldest. Therefore, the probability of this happening is 1 in 120 or 0.83%.
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Answer:
98
Step-by-step explanation:
So we have the two functions:
And we want to find the value of r(g(4)).
To do so, first find the value of g(4):
Multiply:
Subtract:
Now, substitute this into r(g(4)):
And substitute this value into r(x):
Square:
Subtract:
Therefore:
The probability of the fair spinner landing only once on yellow in two spins is 0.375. Option C is the correct option.
<h3>What is probability?</h3>Probability of an event is the ratio of number of favorable outcome to the total number of outcome of that event.
As, the probability of an event can not be more than the number 1.
Thus, the probability of failure of a event is equal to the difference of the 1 to the success of the event. As,
The data in the theoretical probability table for the fair spinner having 4 different color is listed below.
Yellow Frequency
0 9
1 6
2 1
For the 4 different color (red, green, blue, yellow) and rotating the spinner two times, there is total 16 outcomes as,
{RR, BB, YY, GG, RB, BR, RY, YR, RG, GR, BY, YB, BG, GB, YG, GY}
Total number of Y appear probability at one time in the table is 6. Thus the probability of the fair spinner landing only once on yellow in two spins is,
Hence, the probability of the fair spinner landing only once on yellow in two spins is 0.375. Option C is the correct option.
Learn more about the probability here;