How many outcomes are in the sample space for these probability problems?

[franz32]

1. Suppose that two cards are dealt from a standard 52-card poker deck. Let A be the event that the sum of the two cards is 8 (assume that aces have a numerical value of 1). How many outcomes are in A?

Where I got stuck [WIGS]: Are the suits important here? So, there are a lot of outcomes then if repetition of value is permitted... How do I define my "outcomes" here... say 2 of hearts and 6 of diamonds?

2. Consider the experiment of choosing coefficients for the quadratic equation ax^2 + bx + c = 0. Characterize the values of a, b and c associated with the event A: Equation has complex roots.

[WIGS]: I don't get it well in a sense that I had hard time making a "set" out of the problem... I know that complex roots will occur of b^2 - 4ac < 0.
How will I answer it - I mean, I only need to characterize my answer?

3. A probability-minded despot offers a convicted murder a final chance
to gain his release. The prisoner is given 20 chips, 10 white and 10 black. All 20 are to be placed in the two urns, according to the allocation scheme the prisoner wants, provided that each urn has at least 1 chip in it. The executioner will then pick one of the two urns at random, and from that urn, one chip at random. If the chip selected is white, the prisoner's free, otherwise, he "buys the farm". Characterize the sample space describing the prisoner's possible allocation options. (intuitively, which allocation affords the prisoner the greatest chance of survival?)

[WIGS]: Do I have to list all his possible options? If yes, How will I define my elements to be used? If not, then I'm answering that enclosed in ()?
What could be that greatest chance?


-> I have my ideas but I'm not convinced yet on my own because
these doubts make me uncertain.

 

[cookiemonster]

1. Suit only matters in that it distinguishes one card from another. So a 6 of spades and a 2 of diamonds would be different than a 6 of spades and a 2 of hearts.

2. I'd describe the region by making a 3d graph. Let a = x, b = y, c = z and graph the region described by the inequality b^2 - 4ac < 0.

3. Consider the extreme cases. 1 chip in one pot and 19 in the other and 10 chips in each pot. It'll be one of the two. Calculate the probability of not buying the farm for each.

cookiemonster

 

[M98Ranger]

1. In this situation, the suits are not important as long as the numerical values of the cards are taken into account. This means that a 2 of hearts and 6 of diamonds would be considered the same as a 2 of spades and 6 of clubs. Therefore, when determining the outcomes in event A, you can focus solely on the numerical values of the cards.

To find the number of outcomes in A, you can list all the possible combinations of two cards that add up to 8. These combinations are: 2 and 6, 3 and 5, 4 and 4, 5 and 3, and 6 and 2. Since there are four suits for each numerical value, there are 4 possible outcomes for each combination. Therefore, there are a total of 20 outcomes in A.

2. In this experiment, the values of a, b, and c represent the coefficients of the quadratic equation ax^2 + bx + c = 0. For the equation to have complex roots, the discriminant (b^2 - 4ac) must be less than 0. This means that the values of a, b, and c must be chosen in a way that satisfies this condition.

To characterize the values of a, b, and c associated with event A, you can say that a and c can be any value, while b must be chosen in a way that makes the discriminant negative. For example, a = 1, b = 2, and c = 3 would satisfy this condition, as the discriminant would be -8.

3. The sample space in this scenario would consist of all the possible ways the prisoner can allocate the 20 chips between the two urns. This includes options such as placing 10 white chips in one urn and 10 black chips in the other, or placing 5 white and 5 black chips in each urn, or any other possible combination as long as each urn has at least 1 chip.

Intuitively, the allocation that affords the prisoner the greatest chance of survival would be to evenly distribute the chips between the two urns. This means placing 10 white and 10 black chips in each urn. This way, no matter which urn the executioner chooses, there is a 50% chance of picking a white chip and the prisoner's chances of survival are maximized.