Help! Probability... (1 Viewer)

[sAppY..]

Probability is strangling me, could anyone hep me with the following 2 qs?

1. two tennis players are said to have a probabilitiy of 2/5 and 3/4 respectively of winning a tournament. Find the probability that
a) 1 of them will win
b) neither one will win

2. In a bath of 100 cars, past experiene would suggest that 3 could be faulty. If 3 cars are selected at random, find the probability that
a)1 is faulty

 

[pLuvia]

1. a)
P(1 of them will win)=P(first will win, other wont or second will win, other wont)
=2/5*1/4+3/5*3/4
=11/20

b) P(none will win)
=3/5*1/4
=3/20

2.
a) P(faulty car)=3/100
P(1 is faulty)=3C1*(3/100)*(97/100)^2
=0.0847 (4 dp)

 

[sAppY..]

Amazing! Got the first Q. Thanks a lot.
For second Q, dont realli understand.
If we draw out one false car (i.e. 3/100), isn't left us with 99 cars, and (3/100 x 97/99 x 96/99) ?

 

[jake2.0]

You select the 3 cars at once.

 

[pLuvia]

They would specify if the cars were selected without replacement. I don't think combinatorics is taught in 2u, but another way of doing it which uses the same logic as combinatorial probability

Since we assume the cars are chosen all at once

The faulty could be the first of the three, second of the three or third of the three so:
3/100*97/100*97/100+97/100*3/100*97/100+97/100*97/100*3/100
=0.0847

 

[lolokay]

Amazing! Got the first Q. Thanks a lot.
For second Q, dont realli understand.
If we draw out one false car (i.e. 3/100), isn't left us with 99 cars, and (3/100 x 97/99 x 96/99) ?

the question says past experience suggests 3 of 100 will be faulty. This means the equivalent of saying 1 car has 3/100 chance of being faulty, not that given 100 cars exactly 3 will be faulty.