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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 will always hit the target the probability of hitting an odd number three times i class=

Respuesta :

absor201

Answer:

3.37 times

Step-by-step 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 !

Answer:

the answer is 2.25

Step-by-step 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

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