course Mth 152 FGyϐ{Ԝassignment #008 008. Conditional probabilities, more probabilities Liberal Arts Mathematics II 02-17-2008
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20:11:40 `q001. Note that there are 7 questions in this assignment. 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. What is the probability that the card is the Jack or the Queen of Diamonds?
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RESPONSE --> There are 6 red face cards. The probability that the card is either the Jack or Queen of Diamonds would be 2/6 or 1/3. confidence assessment: 2
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20:11:50 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.
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RESPONSE --> Correct answer. self critique assessment: 3
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20:12:47 `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?
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RESPONSE --> Since the first card dealt was a face card, there are 11 face cards remaining out of 51 cards. The probability that the next card dealt will be a face card is 11/51. confidence assessment: 2
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20:13:21 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.
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RESPONSE --> Correct answer. self critique assessment: 3
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20:18:19 `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?
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RESPONSE --> The probability that a five-flip sequence will result in exactly four heads is 4/16 or 1/4. confidence assessment: 1
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20:19:00 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.
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RESPONSE --> Correct answer. self critique assessment: 3
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20:25:54 `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?
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RESPONSE --> There are 18 possibilities with the first die being even. Only four of these possibilities would result in a total greater than 9. So the probability that the total on the two dice will be greater than 9 is 4/18 or 2/9. With unconditional probability, there are 36 possible outcomes. Of these possible outcomes, 6 outcomes would result in a total greater than 9. 6/36 or 1/6. confidence assessment: 1
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20:33:07 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.
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RESPONSE --> Correct on the first part of the problem, but I am confused about the second part of the problem. If there are no conditions, then there would be six possible outcomes greater than 9. self critique assessment: 1
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20:37:29 `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?
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RESPONSE --> There are two possibilities for the first spin (3 and 5). When combining the two spins, there are a total of 10 outcomes. Six of these combinations would result in an odd number. P = 6/10 or 3/5 confidence assessment: 1
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20:39:12 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.
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RESPONSE --> I was thinking we were looking for an odd result, not an even result. I understand how to find the result. self critique assessment: 2
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20:43:03 `q006. What is the probability that two consecutive cards dealt (without replacement) from a full deck will both be Hearts?
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RESPONSE --> There are 13 hearts in a deck of cards. The probability that the first card will be a heart would be 13/52. There would be 12 remaining hearts out of 51 cards. 13/52 * 12/51 = 156/2652 or 26/427 confidence assessment: 1
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20:46:08 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.
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RESPONSE --> I was not thinking about the two possible orders for any 2-card hand. I was on the right track. self critique assessment: 2
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21:05:55 `q007. What is the probability that two consecutive cards dealt (without replacement) from a full deck will both be of the same suit?
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RESPONSE --> 52 cards, 4 suits, 13 cards in each suit: After the first card is dealt, there would be 51 cards remaining and 12 cards of the suit in your hand. The probability of the second card being of the same suit would be 12/51. confidence assessment: 0
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21:06:41 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.
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RESPONSE --> Correct answer, but I didn't reduce it. self critique assessment: 2
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