we have a square on an orthanomal plain and its divided in to 3 section red/orange/yellow

Question

Adam and Ali are playing a game of darts,each person gets 2 throw at the square,the probability of hitting the square is 60%,yellow is 1/4,orange is 5/12 and red is 1/3

and you get 1 point for hitting yellow 

2 point for orange 

3 point for red 

what is the probability of getting at most 1 point?

what is the probability of getting 3 points?

http://imgur.com/a/KHwZn

Answer 1

Let's break down some of the probabilities first:

The probability of missing P_miss = 40%
The probability of hitting yellow P_y = 60% * (1/4) = 15%
The probability of hitting orange P_o = 60% * (5/12) = 25%
The probability of hitting red P_r = 60% * (1/3) = 20%

Let's double check that we have everything: 40% + 15% + 25% + 20% = 100%

 

First, we need to find the probability of getting at most 1 point. To get that, we need to add the probability of missing twice and the probability of missing once and hitting yellow once. The probability of missing twice is P_miss * P_miss. The probability of missing then hitting yellow is P_miss * P_y. The probability of hitting yellow then missing is P_y * P_miss. All in all we get:

 

P_miss * P_miss + P_miss * P_y + P_y * P_miss

= P_miss² + 2 * P_miss * P_y

= (.40)² + 2 * (.40) * (.15)

= .16 + .12

= .28

= 28%

 

Now we need to find the probability of getting exactly 3 points. That can be done in several ways: miss then hit red, hit red then miss, hit yellow then hit orange, or hit orange and then hit yellow. The equation for this combined probability is:

 

2 ( P_miss * P_r ) + 2 ( P_y * P_o)

= 2 ( .20 * .20 ) + 2 ( .15 * .25 )

= .08 + .075

= .155

= 15.5%