#$&* course Mth 152 November 18 - 2:03am Question: `q001. Suppose that a card is dealt from a well-shuffled deck, and that you can tell by the reflection in your opponent's reading glasses that the card is a red face card. However you can't tell any more than that.
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Given Solution: In this case your knowledge that the card is a red face card limits the possibilities to six: The Jack of Hearts or Diamonds, the Queen of Hearts or Diamonds, or the King of Hearts or Diamonds. The probability that the card is one of the two specified cards is therefore 2 / 6 = 1/3. Note that without any limits on the possibilities, the probability that the card is the Jack or Queen of Diamonds is only 2 / 52 = 1 / 26. Note also that the probability that a card is a red face card is 6 / 52 = 3/26. If we divide the first probability by the second we get 1/26 / ( 3/26) = 1/26 * 26/3 = 1/3. Thus the probability that a card is the Jack or Queen of Diamonds, given that it is a red face card, is equal to the probability that it is the Jack or Queen of Diamonds (and a face card), divided by the probability that it is a red face card. This statement has the form 'The probability of B, given A, is equal to the probability of A ^ B divided by the probability of A'. This statement is abbreviated to the form P(B | A) = P(A ^ B) / P(A). This is the formula for Conditional Probability. In this problem the outcome was Jack or Queen of Diamonds, and the condition was that we have a red face card. Self-critique: It all falls down to 1/3 but I should have worked in the probability of the whole 52-card deck. ------------------------------------------------ Self-critique rating: 2 ********************************************* Question: `q002. Suppose that a face card is the first card dealt from a full deck of well-shuffled cards. What is the probability that the next card dealt (without replacement) will also be a face card? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A face card in a deck is a person card. There are 12 face cards in a full deck - Jack, Queen, and King. After choosing one we have 11 possible choices and after choosing a second, we have 10, and so on. As the number of choices goes down, so does the amount of cards in the deck - leaving us with the probability of 11/51, 10/50, 9/49… etc. Since we are asking for the first card turnover though, the probability is 11/51. Which in formula is: P (B | A) = P (A ^ B) / P (A) = [12 / 52 * 11 / 51] / [12 / 52] = 11 / 51 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We know that after the first card is dealt there are 11 face cards left out of the original 12, and 51 cards left in the deck. The probability is therefore obviously 11/51. We can also analyze this situation as a conditional probability. B stands for 'a face card is dealt on the second card' while A stands for 'a face card is dealt on the first card'. So the event A ^ B stands for 'a face card is dealt on the first card and on the second', with probability 12/52 * 11/51. A stands for 'a face card is dealt on the first card', with probability 12 / 52. So P(B | A) stands for 'a face card is deal on the second card given that a face card is dealt on the first'. By the formula we have P(B | A) = P ( A ^ B ) / P(A) = [ 12 / 52 * 11 / 51 ] / [ 12 / 52 ] = 11 / 51, which of course we already knew from direct analysis. Self-critique: OKAY ------------------------------------------------ Self-critique rating: OKAY ********************************************* Question: `q003. Given that the first clip of a coin is Heads, what is the probability that a five-flip sequence will result in exactly four Heads? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Same as before but with coin flips. We flip the coin five times, with a possibility of heads of tails each time. This makes 2 possibilities for each of the 5 flips = 2^5 = 32 outcomes. Half of the flips have a possibility of being heads and half tails which equal 16. Out of those 16, there are four times when the five flip sequence can result in exactly four Heads - leaving us with a ¼ probability. Which in formula is: P (B | A) = P (A ^ B) / P (A) = (4/32) / (2/1) = (4 / 32) * (2 / 1) = 4 / 16 = 1/4 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If we were to list the 2^5 = 32 possible outcomes for five flips, we would find that 16 of them have 'heads' on the first flip, and that of these 16 there are 4 outcomes with exactly four 'heads'. The probability therefore looks like 4 / 16 = 1/4, which is correct. To verify this by the formula P( B | A ) = P( A ^ B) / P(A), we let B stand for the desired event of exactly four 'heads' and A for the 'given' event of 'heads' on the first flip. On five flips, P(A) = 16 / 32 = 1/2 (probability of 'heads' on the first flip), which P(B ^ A) = 4 / 32 (four of the 32 possible outcomes have 'heads' on the first flip and exactly four 'heads'). The formula therefore gives us P( B | A ) = P( A ^ B) / P(A) = (4/32) / (2/1) = (4 / 32) * (2 / 1) = 4 / 16 = 1/4. Self-critique: OKAY ------------------------------------------------ Self-critique rating: OKAY ********************************************* Question: `q004. Given that the first of two dice comes up even, what is the probability that the total on the two dice will be greater than 9? How does this compare with the unconditional probability that the total of two fair dice will be greater than 9? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We are looking at two dice coming up as an even number but the amount of the two numbers has to be greater than 9. We are looking for the probability and the unconditional probability. The best way to find the numbered pairs is to make an example of the sample space. We must take into account that the dice only rolls up to 6. Our options then become: (2,1)(2,2)(2,3)(2,4)(2,5)(2,6) (4,1)(4,2)(4,3)(4,4)(4,5)(4,6) (6,1)(6,2)(6,3)(6,4)(6,5)(6,6) There are six options in each of the three rows making 18 overall pairs. Of these 18 pairs (4,6)(6,4)(6,5)(6,6) make a combined number greater than 9. This makes our probability 4/18, simplifying down to 2/9. Which in formula is: P (B | A) = P (A ^ B) / P (A) = (1/9) / (1/2) = (1/9) * (2/1) = 2/9 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We can list the sample space of dice possibilities for which the first number is even. The sample space is { (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }. We note that there are 18 elements in the sample space. We then find the corresponding totals, which are 3, 4, 5, 6, 7, 8 5, 6, 7, 8, 9, 10 7, 8, 9, 10, 11, 12. Of these 18 totals, 4 are greater than 9. Thus the probability that the total of two dice will be greater than 9, given that the first is even, is 4/18 = 2/9. To verify this by the formula P( B | A ) = P( A ^ B) / P(A), we let B stand for the set of all dice pairs which give a total greater than 9, and A for the set of all dice pairs where the first die shows an even number. We have seen that A = { (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }. Listing the elements in B we find that B = { (4, 6), (6, 4), (5, 5), (6, 5), (5, 6), (6, 6) }. There are 6 elements in this set. A ^ B consists of the set of elements common to both A and B, or { (4, 6), (6, 4), (6,5), (6, 6) }. Since there are 4 elements in A ^ B, 18 elements in A, and 36 elements in the sample space for two dice, it follows that P(A) = 18 / 36 = 1/2 and P(A ^ B) = 4 / 36 = 1/9. Therefore the probability we are looking for, P(B | A), is given by P(B | A) = P(A ^ B) / P(A) = (1/9) / (1/2) = (1/9) * (2/1) = 2/9. This is in agreement with the previous result obtained by listing. STUDENT QUESTION I understood this but I’m confused about where the 36 examples in the sample space come from in the probability formula? INSTRUCTOR RESPONSE If you list all the possible outcomes for 2 dice, you find that there are 36 possibilities. They can be listed in a table with 6 columns and 6 rows. Self-critique: OKAY ------------------------------------------------ Self-critique rating: OKAY ********************************************* Question: `q005. A spinner has numbers 2, 3, 4, 5 and 6. Given that the first number is odd, what is the probability that the sum of the results on two consecutive spins is even? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A spinner has the numbers 2,3,4,5, and 6. Given that the first number is odd which can be 3 or 5, what is the probability that the sum of the results on two consecutive spins is even. First we must look at the results of all the spins beginning with an odd number or a sample space = (3,2)(3,3)(3,4)(3,5)(3,6)/(5,2)(5,3)(5,4)(5,5)(5,6). Of these 10 spun pairs, 4 of them come out as an even numbers: (3,3)(3,5)(5,3)(5,5). The probability of the set is 4/10, simplifying to 2/5. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The set of possibilities for which the first number is odd is { (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) }. There are therefore 10 possibilities. Of these 4 add up to an even total, so the probability that the total is even, given that the first number is odd, is Probability of B given A = 4/10 = 2/5. To verify this by the formula P( B | A ) = P( A ^ B) / P(A), we let B stand for the set of all pairs that add up to an even number and A for the set of all pairs for which the first number is even. The sample space for two spins has 5 * 5 = 25 elements. Of these, only the four outcomes (3, 3), (3, 5), (5, 3) and (5, 5) for which both spinners land on odd numbers are in the set A ^ B. Thus P(A | B) = 4/25. The set A consists of the 10 pairs listed earlier. So P(A) = 10/25 = 2/5. Thus P(B | A) = P(A ^ B) / P(A) = (4/25) / (2/5) = (4/25) * (5/2) = 2/5 in agreement with our previous result. Self-critique: OKAY ------------------------------------------------ Self-critique rating: OKAY ********************************************* Question: `q006. What is the probability that two consecutive cards dealt (without replacement) from a full deck will both be Hearts? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Two consecutive cards from a 52-card deck with the probability of drawing two hearts from the 13 in the 52-card deck. When we draw the first card with are drawing from the possibility of 13/52. When we draw the second card we are drawing from the possibility of 12/51. And so on, but in this case we are only looking at the first two consecutive cards drawn. To find any two-card hand we use the formula C (52,2). The possibility of any 2 hands is 1,326. But to find a two card hand out of the 13 hearts we use C (13,2) = 156/2 = 78. The overall probability of drawing two consecutive hearts is therefore 78/1,326. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: When the first card is dealt there are 13 Hearts in a deck of 52. When the second card is dealt there are 12 Hearts left in the remaining 51 cards. The order in which the cards are dealt does not matter, and there are two possible orders for any 2-card 'hand'. The number of ways to get 2 Hearts is therefore C(13,2) = 13 * 12 / 2 = 156 / 2 = 78. The number of possible 2-card 'hands' is C(52, 2) = 52 * 51 / 2 = 1326. The probability of obtaining two Hearts is therefore 78 / 1326, which can be reduced or expressed as a decimal. STUDENT COMMENT The part I find trickiest is to remember to divide the 13*12 by 2 and the 52*51 by 2 because I seem to forget to do so. INSTRUCTOR RESPONSE Remember that in this case the order in which the cards are dealt doesn't matter; the player can rearrange them in any way he or she chooses. You therefore need to divide by the number of possible orders. Self-critique: OKAY ------------------------------------------------ Self-critique rating: OKAY ********************************************* Question: `q007. What is the probability that two consecutive cards dealt (without replacement) from a full deck will both be of the same suit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Similar to the question above but we are looking for two consecutive cards of any of the same suit instead of just hearts. The first card we draw can be any card from any suit, which contains 13 cards. The second card drawn has to be one of the remaining 12 cards from the same suit then. The probability of drawing a card from the same suit us 12/51 which can be simplified down to 4/1 when divided by 3. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A quick common-sense solution tells us that sense the first card can be anything, then since of the 51 remaining cards, there remain remain 12 cards that match the suit of the first the probability must be 12/51 = 4/17. STUDENT QUESTION For the previous problem, we used two hearts, so how come this problem isn’t completed the same way that the last one was, given that we are finding 2 of the same suit? INSTRUCTOR RESPONSE In the preceding problem the suit was specified. The condition couldn't be satisfied unless the first card was a heart. In this problem any first card is OK; the only condition is that the second card be of the same suit. So the probability here is significantly higher. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q007. What is the probability that two consecutive cards dealt (without replacement) from a full deck will both be of the same suit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Similar to the question above but we are looking for two consecutive cards of any of the same suit instead of just hearts. The first card we draw can be any card from any suit, which contains 13 cards. The second card drawn has to be one of the remaining 12 cards from the same suit then. The probability of drawing a card from the same suit us 12/51 which can be simplified down to 4/1 when divided by 3. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A quick common-sense solution tells us that sense the first card can be anything, then since of the 51 remaining cards, there remain remain 12 cards that match the suit of the first the probability must be 12/51 = 4/17. STUDENT QUESTION For the previous problem, we used two hearts, so how come this problem isn’t completed the same way that the last one was, given that we are finding 2 of the same suit? INSTRUCTOR RESPONSE In the preceding problem the suit was specified. The condition couldn't be satisfied unless the first card was a heart. In this problem any first card is OK; the only condition is that the second card be of the same suit. So the probability here is significantly higher. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!