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german
3 years ago
6

If the racetrack publishes that the odds in favor of a horse winning a race are 5 to 9, what is probability that the horse will

not win the race?
Mathematics
2 answers:
crimeas [40]3 years ago
5 0
I divided 5 by 9 and multiplied it by 100 to get the percentage for the probability the horse will win. I got 55.55%. I subtracted 55.55 from 100 and got 44.45% chance of not winning the race.
LUCKY_DIMON [66]3 years ago
3 0

Answer: \dfrac{9}{14}

Step-by-step explanation:

We know that , In probability the odds for an event is the ratio of the favorable outcomes to the unfavorable outcomes.

We are given that ,

The racetrack publishes that the odds in favor of a horse winning a race are 5 to 9 .

i.e. favorable outcomes for horse winning a race = 5

i.e.  unfavorable outcomes for horse winning a race = 9

Now , Total outcomes = favorable outcomes+  unfavorable outcomes

= 5+9 = 14

Then, the probability that the horse will not win the race=\dfrac{\text{Unfavorable outcomes}}{\text{Total outcomes}}

=\dfrac{9}{14}

Hence, the probability that the horse will not win the race =\dfrac{9}{14}

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a) 0.1423

b) 0.2977

c) 0.56

Step-by-step explanation:

For each driver stopped for speeding, there are only two possible outcomes. Either they have invalid licenses, or they do not. The probability of a driver having an invalid license is independent from other drivers. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

13 percent of the drivers stopped for speeding have invalid licenses.

This means that p = 0.13

14 drivers are stopped

This means that n = 14

(a) None will have an invalid license.

This is P(X = 0)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{14,0}.(0.13)^{0}.(0.87)^{14} = 0.1423

(b) Exactly one will have an invalid license.

This is P(X = 1)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{14,1}.(0.13)^{1}.(0.87)^{13} = 0.2977

(c) At least 2 will have invalid licenses.

Either less than 2 have invalid licenses, or at least 2 does. The sum of the probabilities of these events is decimal 1. Mathematically, this is

P(X < 2) + P(X \geq 2) = 1

We want P(X \geq 2)

So

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1) = 0.1423 + 0.2977 = 0.44

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.44 = 0.56

8 0
3 years ago
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