# Need help with probability ...Please ....!

• Jun 22nd 2005, 05:35 AM
rockwell
Need help with probability ...Please ....!
The archery tournament:

At an archery tournament, eight of the twenty competitors were left-handed and the remainders were right-handed.

1. Find the probability that an archer chosen at random is left-handed.
2. If a pair of archers is chosen at random is left-handed. Find the probability that exactly one of the pair is left-handed.

Due to the weather conditions it is more difficult for left-handers to hit the target. Practice sessions indicated that the probability of a left-hander hitting the target was 0.7, while for right-hander it was 0.9.
3. Find the probability that an archer chosen at random is left-handed and hits the target.
4. Find the probability that an archer chosen at random hits the target.
5. If an archer hits the target, find the probability that the person was right-handed.
6. Fin the probability that two archers chosen at random are the same hand and both hit the target.
7. Draw up a table which shows the probability for each outcome possible for two randomly chosen arches.
• Jun 22nd 2005, 07:43 AM
hpe
Quote:

Originally Posted by rockwell
At an archery tournament, eight of the twenty competitors were left-handed and the remainders were right-handed.

1. Find the probability that an archer chosen at random is left-handed.

Obviously 8 out of 20 or 40%.
Quote:

2. If a pair of archers is chosen at random is left-handed. Find the probability that exactly one of the pair is left-handed.
I don't quite understand. Do you mean "if at least one archer in a randomly chosen pair is left-handed"? In that case the solution is as follows:
There are 20*19/2 = 190 possible pairs. Of these, 12*11/2 = 66 pairs consist only of right-handed archers, 8*7/2 = 28 consist only of left-handed archers, 190-66 = 124 have at least one left-handed archer, and 190 - 66 - 28 = 96 = 8*12 consist of a left-handed and a right-handed archer. So the probability is 96/124 = 24/31.

Let me know if this helps, then we'll get to the rest. Question 3 is done with the multiplication rule for conditional probabilities, 4 with total probability, 5 with Bayes rule.
• Jun 23rd 2005, 04:26 AM
rockwell
Yes...thanks you .....but i don't understand anything about probability ...so can u tell me how to do question 3,4,5,6,7 ...because my brother ...ask me for help ....but i studied for a long time a go ....i don't remmember....so can u help me ...solve this ...thanks you ...
• Jun 23rd 2005, 07:36 AM
hpe
Quote:

Originally Posted by rockwell
Yes...thanks you .....but i don't understand anything about probability ...so can u tell me how to do question 3,4,5,6,7 ...because my brother ...ask me for help ....but i studied for a long time a go ....i don't remmember....so can u help me ...solve this ...thanks you ...

Get your brother to post the questions - or do your homework yourself. :D