A probability problem by A Former Brilliant Member

Probability Level pending

A box contains 12 identical balls. Five are brown and seven are not. Two balls are drawn together. What is the probability that at least one brown ball is selected?

7 22 \frac{7}{22} 25 144 \frac{25}{144} 5 33 \frac{5}{33} 15 22 \frac{15}{22}

A Former Brilliant Member by A Former Brilliant Member

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

The balls selected can be 1 brown ball and 1 not brown ball or 2 brown balls.

P P = = ( 5 C 1 ) ( 7 C 1 ) 12 C 2 \frac{(5C1)(7C1)}{12C2} + + 5 C 2 12 C 2 \frac{5C2}{12C2} = = 35 66 \frac{35}{66} + + 10 66 \frac{10}{66} = = 15 22 \frac{15}{22}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...