Eric has a bag of candy. There are 6 grape pieces, 5 orange pieces and 3 lime pieces. Eric picks one piece of candy at random and eats it. Then he takes a second piece. What is the probability that both were grape?
15/91
7/50
7/9
3/20
2007-12-29 03:46:48 · 4 answers · asked by sunny p 1 in Science & Mathematics ➔ Mathematics
Probability that the first was grape = 6 /14 = 3/7.
Since he ate the first piece, the probability that
the second is grape is 5/13.
Probability that both are grape = 3/7 * 5/13 = 15/91.
2007-12-29 03:54:48 · answer #1 · answered by steiner1745 7 · 0⤊ 0⤋
Total = 14 pieces of candy
prob. of first grape = 6/14
prob. of second grape = 5/13
6/14*5/13=30/182
30/182=15/91
2007-12-29 11:52:20 · answer #2 · answered by Tommy 2 · 1⤊ 0⤋
6+5+3= 14pieces in total (6grape)
prob that first piece is grape= 6/14
now there are 13 pieces left (5grape)
prob that second piece is grape= 5/13
prob that both were grape= (6/14)*(5/13)= 15/91
2007-12-29 11:52:55 · answer #3 · answered by Just me 5 · 1⤊ 0⤋
it would be 6/14+5/13 i think
2007-12-29 11:48:55 · answer #4 · answered by Anonymous · 0⤊ 3⤋