Some basic probability issues. (a) Three dice are thrown. What is the probabilit

Question

Some basic probability issues.

(a) Three dice are thrown. What is the probability that the combined value is 7? What is the most probable value for the sum of the three values?

(b) Two cards are drawn randomly from a standard deck of cards. What is the probability that they have the same denomination?

(c) Three red balls, three blue balls and 3 green balls are placed in an opaque bag. Three balls are drawn at random from the bag. What is the probability that the three balls are all of a different color?

(d) A classical particle is placed in a closed rectangular box that is shaken and placed on a horizontal surface. What is the probability that the particle is in the front half of the box? Explain your reasoning?

(e) A classical particle is placed in a closed two-dimensional rectangular box which has dimensions 3 cm by 5 cm. What is the probability that the particle is within 1 cm of one of the walls? Explain your reasoning?

Explanation / Answer

(a) If we look at two dice we can make a 6 by 6 grid. The third die will have equal probability (1/6) for each number,
Third die other two
6 1 , can't happen
5 2
4 3
3 4
2 5
1 6
Now let us look at the odds of the other two adding to 2, there is only one way this can happen, cat eyes, so 1/36.
There are two ways the other two can add to three (2 and 1 or 1 and 2) so 2/36.
There are three ways the other two can add to four (1 and 3, 2 and 2, and 3 and 1) so 3/36
There are four ways the other two can add to five (1,4; 2,3; 3,2; 4,1) so 4/36
There are five ways the other two can add to six (1,5; 2,4; 3,3; 4,2; and 5,1) so 5/36
These are all mutually exclusive so we can add the probabilities.
The first is the chance that the other two add to 1, which is impossible, so the probability is zero
1/6( zero) + 1/6(1/36) + 1/6(2/36) + 1/6(3/36) +1/6(4/36) + 1/6(5/36) = 1/6(15/36)
=3(5)/6^3 = 5/72
So the chance of getting a total of 6 is 5/72 or 6.9%
the most probable is the average of each added
Average of 1, 2, 3, 4, 5, and 6 is 3 1/2
So, two dice most probable is 3 1/2 + 3 1/2 = 7
So, three dice, the most probable is 3 1/2 + 3 1/2 + 3 1/2 = 10 1/2. However, you never get a ten and a half when you roll three dice. This means that the most probable is a tie between 10 and 11. The chance for each is 1/8 (I did this on a spreadsheet)

(b) We remove the jokers and have 52 cards. There are four of each denomination.
We have a 100% chance of picking our first card. No matter what we picked there are three left in the deck with that same denomination and 51 cards left in the deck
3(17) = 51, so we have a 1/17 chance to pick a matching card. The probability is 1/17.

(c) Again we have a 100% chance of picking one ball, For the second there are six balls that keep us alive and a total of eight balls, so 6/8. For our (crucial) third pick we need to select one of the three balls of the third color and there are seven left in the bag. All of these need to happen, so we multiply the probabilities
1(6/8)(5/7) = 15/28 So a little better than 50:50

(d) Symmetry of the problem shows 50%. We could do experiment, detect the particle, and then flip a coin to decide what we call front and what we call back of the box.

(e) Barring van der Waal's forces and induced dipoles, which would favor the edges we will have area/ total area. Total area is 15 cm^2. Banned area is 1 cm by 3 cm in central region. Successful area is 15-3 = 11 cm^2
Probability of success is 11/15.

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