This is a lot harder then every one that answered my last question thought. Here is the Question!
A die is rolled and a coin is tossed. The die and the coin are fair. What is the probability that the outcome of the roll is a number more than 4 and the outcome of the coin toss is heads? Write your answer as a fraction in lowest terms.
No, im not a Retard.. this is harder than it looks....
2007-05-30 12:41:21 · 7 answers · asked by Thegrinch57 1 in Education & Reference ➔ Homework Help
more than 4 would be 5 and 6
so 2/6 or 1/3
heads 1out of 2
multiply
1/3*1/2=
1/6
2007-05-30 12:45:48 · answer #1 · answered by Kobie D 3 · 0⤊ 0⤋
Well, the chance of a heads is 1/2.
The chance that the number on the die is more than 4 (or in other words, is either a 5 or a 6) is 2/6 or 1/3.
Thus, 1/2 times 1/3 equals 1/6.
2007-05-30 19:44:50 · answer #2 · answered by RandomNormality 3 · 0⤊ 0⤋
For the die, probability that it is more than 4 (i.e., a 5 or 6) is 2/6, or 1/3.
For the coin, probability that it is heads is 1/2.
Then, you just multiply the two probabilities. 1/3 * 1/2 = 1/6
2007-05-30 19:46:35 · answer #3 · answered by derek1079 5 · 0⤊ 0⤋
Since these are independent, just multiply the probabilities.
P(Die is 5 or 6) = 2/6 = 1/3
P(Heads) = 1/2
P(Die is 5 or 6, AND Heads) = (1/3) * (1/2) = 1/6
2007-05-30 19:44:42 · answer #4 · answered by atomicjohnson 3 · 0⤊ 0⤋
1/6
2007-05-30 19:45:39 · answer #5 · answered by Ashley M 1 · 0⤊ 0⤋
die 2/6 = 1/3
coin 1/2
That's all the can help you with. I hope it's right.
2007-05-30 19:47:04 · answer #6 · answered by T T 2 · 0⤊ 0⤋
die- 1/3
coin- 1/2
2007-05-30 19:58:12 · answer #7 · answered by someone 2 · 0⤊ 0⤋