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HELP. ME!!!

A player gets to throw 4 darts at the target shown. Assuming the player will always hit the target, the probability of hitting an odd number three times is times more than the probability of hitting an even number three times.

HELP. ME!!! A player gets to throw 4 darts at the target shown. Assuming the player-example-1
User Enchilada
by
5.4k points

2 Answers

3 votes

Answer:

the answer is 2.25

Explanation:

use the binomial theorem

n = number of trials

r = number of successes

p = probability

* = times

(n! / r! (n-r)!) * (p)^r * (1-p)^(n-r)

Plug everything in for odds with a probability of .6

Plug everything in for evens with a probability of .4

Divide the probability of the odds hitting 3 times by the probability of the evens hitting 3 times

That should get you 2.25

User KonstantinosKokos
by
5.3k points
4 votes

Answer:

3.37 times

Explanation:

Probability of hitting and odd number= 3/5

Probability of hitting and odd number=2/5

Probability of hitting and odd number three times= (3/5)^3

Probability of hitting and odd number three times=(2/5)^3

Now

(3/5)^3 / (2/5)^3 = 3.37

Hence Probability of hitting an odd number three times is 3.37 times more than the probability of hitting an even number three times !

User Janaka Chathuranga
by
4.8k points
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