1. ## Probability question help, please tell how you did it.

Hey, I was just wondering if any one could help me with this question. I already know the answers and have provided them for you, however i would like to know how the answer is obtained.

Q. Football Strategies: A particular football team is known to run 30% of its plays to the left and 70% to the right.A linebacker on the opposing team notes that the right guard shifts his stance most of the time (80%) when plays go to the right and that he uses a balanced stance the remainder of the time. When plays go to the left, the guard takes a balanced stance 90% of the time and the shift stance the remaining 10%. On a particular play, the linebacker notes that the guard takes a balanced stance.

(a) What is the probability that the play will go to the left? Answer:.6585

(b) What is the probability the play will go to the right?

I have spent ages trying to figure it out, so any help will be greatly appreciated.

Cheers,
RaVS.

2. okay we already know that 1) he has balanced stance 2) he will go left or right.
1=a(P(left)+P(right))
1=a((0.9*0.3)+(0.2*0.7)) gives us a=100/41.
a*P(left) = (100/41)*(0.9*0.3) = 27/41 ≈ 0.6585
a*P(right) = 1-0.6585 = 14/41 ≈ 0.3415

I hope that this answer will make sense to you!

3. Hello, RaVS!

I assume you know Bayes' Theorem: . $P(A\,|\,B) \;=\;\frac{P(A \wedge B)}{P(B)}$

A football team is known to run 30% of its plays to the left and 70% to the right.

A linebacker on the opposing team notes that:

When the play goes to the right, the right guard shifts his stance 80% of the time
. . and uses a balanced stance 20% of the time.

When plays go to the left, the guard shifts his stance 10% of the time
. . and uses a balanced stance 90% of the time.

On a particular play, the linebacker notes that the guard takes a balanced stance.

(a) What is the probability that the play will go to the left? . (Answer: 0.6585)

(b) What is the probability the play will go to the right? . (Answer: 0.3415)

(a) We want: . $P(\text{play goes Left }|\text{ balanced stance}) \;=\;\frac{P(\text{play goes Left }\wedge\text{ balanced stance})}{P(\text{balanced stance})}$

The numerator is:
. . $P(\text{play goes Left}\, \wedge\,\text{balanced stance}) \;=\;(0.30)(0.90) \:=\:0.27$

The denominator is:
. . $P(\text{balanced stance}) \;=\;\underbrace{P(\text{Left}\wedge\text{balanced })}_{0.27} + \underbrace{P(\text{Right}\wedge\text{balanced})}_ {(0.70)(0.20)} \;=\;0.41$

Therefore: . $P(\text{play goes Left }|\text{ balanced stance}) \;=\;\frac{0.27}{0.41} \;=\;0.6585365685 \;\approx\;0.6585$

(b) This is the opposite of part (a).

Therefore: . $P(\text{play goes Right }|\text{ balanced stance}) \;=\;1 - 0.6585 \;=\;0.3415$

4. I do know about Baye's theorem but am not sure exactly how it works, could you please explain it in better detail to me?

Thank you to Coscos and Sorban, your help is greatly appreciated.