On a tv game show there are 9 squares. Behind 3 of the squares there is a cash amount to be won. Behind every other square there is a prize. 2 of the squares have already been turned over, one prize and one cash. What is the probability of turning over a cash amount next?
2007-08-05 20:43:34 · 4 answers · asked by chavolf 1 in Science & Mathematics ➔ Mathematics
2/7
There are a total of 9 squares.
Two squares have already been turned over, leaving 7 squares left. Each square left has a 1/7th of a chance of being picked. There are two cash prizes left, so that makes it 2/7th's of a chance of a cash prize being picked.
2007-08-05 20:56:30 · answer #1 · answered by Sparks 6 · 0⤊ 0⤋
so 9 squares
assuming the 1st, 3rd, 5th, 7th, and 9th squaws have prizes
that leaves 4 squares 3 with cash and 1 blank.
1 cash & 1 prize revealed leaving 7 squares
1 blank, 2 cash, and 4 prizes
prob of cash = 2/7 2in 7
if the sequence of prizes starts with the 2nd square then there are 4 prizes and 2 blanks along with 3 cash but the chances for a cash square after 2 turn overs remains 2/7.
2007-08-06 04:58:48 · answer #2 · answered by 037 G 6 · 0⤊ 0⤋
2/7
2007-08-06 03:47:07 · answer #3 · answered by Tim 2 · 0⤊ 0⤋
(3 - 1) / ([3-1] + [6 - 1])
2 / (2 + 5)
2 / 7
answer: 2:5 ratio or 2/7 or 28 4/7%
2007-08-10 02:38:48 · answer #4 · answered by Jun Agruda 7 · 2⤊ 0⤋