#$&* course Mth 152 6/23 6:00pm If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: In this case your knowledge that the card is a red face card limits the possibilities to six: The Jack of Hearts or Diamonds, the Queen of Hearts or Diamonds, or the King of Hearts or Diamonds. The probability that the card is one of the two specified cards is therefore 2 / 6 = 1/3. Note that without any limits on the possibilities, the probability that the card is the Jack or Queen of Diamonds is only 2 / 52 = 1 / 26. Note also that the probability that a card is a red face card is 6 / 52 = 3/26. If we divide the first probability by the second we get 1/26 / ( 3/26) = 1/26 * 26/3 = 1/3. Thus the probability that a card is the Jack or Queen of Diamonds, given that it is a red face card, is equal to the probability that it is the Jack or Queen of Diamonds (and a face card), divided by the probability that it is a red face card. This statement has the form 'The probability of B, given A, is equal to the probability of A ^ B divided by the probability of A'. This statement is abbreviated to the form P(B | A) = P(A ^ B) / P(A). This is the formula for Conditional Probability. In this problem the outcome was Jack or Queen of Diamonds, and the condition was that we have a red face card. Self-critique: ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q002. Suppose that a face card is the first card dealt from a full deck of well-shuffled cards. What is the probability that the next card dealt (without replacement) will also be a face card? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 11 face cards, 51 cards left in deck so probability is 11/51. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: I understand how this come out, still not understanding the complete formula. ------------------------------------------------ Self-critique rating: 2 ********************************************* 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: 2^5 possible outcomes for five flips which equals out to 32. So you would have 4/16=1/4 probability of a five flip sequence resulting in heads. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If we were to list the 2^5 = 32 possible outcomes for five flips, we would find that 16 of them have 'heads' on the first flip, and that of these 16 there are 4 outcomes with exactly four 'heads'. The probability therefore looks like 4 / 16 = 1/4, which is correct. To verify this by the formula P( B | A ) = P( A ^ B) / P(A), we let B stand for the desired event of exactly four 'heads' and A for the 'given' event of 'heads' on the first flip. On five flips, P(A) = 16 / 32 = 1/2 (probability of 'heads' on the first flip), which P(B ^ A) = 4 / 32 (four of the 32 possible outcomes have 'heads' on the first flip and exactly four 'heads'). The formula therefore gives us P( B | A ) = P( A ^ B) / P(A) = (4/32) / (2/1) = (4 / 32) * (2 / 1) = 4 / 16 = 1/4. Self-critique: ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q004. Given that the first of two dice comes up even, what is the probability that the total on the two dice will be greater than 9? How does this compare with the unconditional probability that the total of two fair dice will be greater than 9? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) 3, 4, 5, 6, 7, 8 5, 6, 7, 8, 9, 10 7, 8, 9, 10, 11, 12 18 totals which 4 are greater than 9. So 4/18=2/9 being the probability of total two dice being greater than 9 and the first two being even. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ------------------------------------------------ Self-critique rating: 3 ********************************************* 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: 2,3,4,5,6-(3,2),(3,3),(3,4),(3,5),(3,6),(5,2),(5,3),(5,4),(5,5),(5,6) 10 possibilities, then 4 numbers will result in even with the first number being odd. So probability would be 4/10=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: I understand how the solution is done but when seeing the given solution I’m not sure if I should be checking mine and putting it with my solution. ------------------------------------------------ Self-critique rating: 2
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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: How is it that you figured 52*51/2=1326
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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. ********************************************* Question: `q008. On a roll of two dice, what is the probability that one of the numbers is a 6, given that the total is 10? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2 dice =6*6=36 possible outcomes, If you have two numbers totaling 10 with chance number being 6. So if you had 2 chances the probability would be 2/10=1/5. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating:2