Posted: 11/17/2015 11:20:39 AM EDT
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Wife was taking a online college entrance test and asked me to look at this question.
As I see it the correct answer is 1/144 but I know they are wanting to see 1/24 Am I correct or am I over thinking it. (1/2×1/6)×(1/2×1/6) vs. 1/2 × 1/2×1/6 
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Quoted:
I believe it is the second answer you posted. You should only count the dice once. (1/2)*(1/2)*(1/6)=1/24 This is correct, they are three seperate and independent events. So (1/2)*(1/2)*(1/6)=1/24. To get the 1/144 answer in the way that OP wrote it out would imply that the coin tosses coming up heads are contingent on the dice coming up 4. |
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Quoted: I think the odds of getting two heads would be 1/3. Possible coin combos: 2 heads 1 heads and 1 tails 2 tails Multiply that by 1/6 th chance of rolling a 4. 1/4--two heads 1/2--one each 1/4--two tails |
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Quoted:
This is right. 1/2 for each coin and 1/6 for the one die, answer is 1/24. Quoted:
Quoted:
I believe it is the second answer you posted. You should only count the dice once. (1/2)*(1/2)*(1/6)=1/24 This is right. 1/2 for each coin and 1/6 for the one die, answer is 1/24. This is the correct answer. For probabilities of multiple outcomes, you multiply the probability of each individual out come. (1/2)*(1/2)*(1/6)=1/24 For the Quarter, probabililty of A happening (heads or tails) is P(A)=X_A/T where X_A is the number of ways A can happen, T is the total number of elementary events. Thus: P(heads)=1/2, and P(tails)=1/2. Same goes for a 6 sided dice (Assuming that it is numbered 1 trough 6 which is missing in the problem statement): P(some_#)=1_# / 6_sides The rule applies: If two events A and B are independent then the probability that both event A and B will occur is equal to the product of their respective probabilities. Therefore, P (A_coin1, and B_coin2, and C_dice)=P(A)*P(B)*P(C) or 1/2*1/2*1/6 = 1/24 |
