Drag tiles to fill in the blanks to complete the expression that could be used to predict the number of winners of a game. Tiles
1 answer:
Answer:
First blank - A)Total Number of Possible Outcomes
Second blank - B)Number of Winners
Step-by-step explanation:
The exact question is as follows :
We know that,
Probability of an event is equals to Total number of Favorable outcomes divided by Total number of outcomes
So,
P(event) = Number of favorable outcomes ÷ Total Number of Possible Outcomes
As we have to predict the Number of winners of a game
So,
P(Number of winner) = Number of winners ÷ Total number of Contestants
∴ we get
P(event) = Number of favorable outcomes ÷ Total Number of Possible Outcomes
= Number of winners ÷ Total number of Contestants
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Answer:
a) The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.
b) The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.
c) The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.
Step-by-step explanation:
For each smoker, there are only two possible outcomes. Either they started smoking before 21 years old, or they did not. Smokers are independent, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
50% of adult smokers started smoking before 21 years old.
This means that
5 smokers 21 years old or older are randomly selected, and the number of smokers who started smoking before 21 is recorded.
This means that .
a) The probability that at least 2 of them started smoking before 21 years of age is
This is:
In which
The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.
b) The probability that at most 4 of them started smoking before 21 years of age is
This is:
In which
The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.
c) The probability that exactly 3 of them started smoking before 21 years of age is
This is P(X = 3). So
The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.