#$&* course Mth 152 7/21/13 939 pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aIn 1-and-1 shooting you only get a second shot if you make the first. So there are 3 possibilities: miss the first, don't get another shot make the first, get another shot and make it make the first, get another shot and miss it prob of 0 = prob of miss on first shot = .3 prob of 1 = prob of hit on first and miss on 2d = .3 * .7 = .21 prob of 2 = prob ot hit * prob of hit = .49. 'Hits' happen with 70% or .7 probability, misses with probability 30% or 3. The theoretical probability of missing the first shot (and therefore not getting another shot, thereby scoring 0) is just probability of miss = .30 The theoretical probability of 1 miss and 1 hit is probability of hit * probability of miss = .7 * .3 = .21 The theoretical probability of 2 hits is probability of hit * probability of hit = .70 * .70 = .49. Note that these probabilities add up to .30 + .21 + .49 = 1, as they must since these three events cover all possibilities. To use the table, randomly pick a starting point. Let numbers 1-7 correspond to making the free throw, with 8, 9 and 0 corresponding to misses. Go down the list, or across the list in an order you decided before looking at the list. Read two digits from the list and see if they correspond to two 'hits', two 'misses' or a 'hit' and a 'miss'. Record your result as 'hit-miss', 'miss-hit', 'miss-miss' and 'hit-hit'. Read two more digits and record your result as 'hit-miss', 'miss-hit', 'miss-miss' and 'hit-hit'. Continue until you have the required number of results. Tally how many times you got 0 'hits', 1 'hit', 2 'hits' etc.. Any outcome that starts with a 'miss' corresponds to zero point. 'Hit-miss' corresopnds to 1 point and 'hit-hit' corresponds to 2 points. Determine the percent of time you got each number of points, and compare to the theoretical probabilities .30, .21 and .49. *&*& ** Self-critique I am not sure if I finished that correctly. ------------------------------------------------ Self-critique Rating:
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Given Solution: `aYour probabilities are given as 1/2, 1/6 and 1/3. These can all be expressed in terms of the common denominator 6: 1/2 = 3/6, 1/6 = 1/6, 1/3 = 2/6. So a move to the right has 3 chances out of 6, a move to the left has 1 chance out of 6 and a move straight has 2 chances out of 6. You can simulate this by letting the three digits 1, 2, 3 stand for a move to the right, the single digit 4 for a move to the left and the two digits 5, 6 for a straight move. The remaining digits 0, 7, 8, 9 don't stand for anything, and if you land on one of these numbers you just move to the next number. So according to your the first two columns of you table, how many times do you move to the right, how many to the left, how many straight and where do you end up? ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: Query Add comments on any surprises or insights you experienced as a result of this assignment. `Self-critique ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: Query Add comments on any surprises or insights you experienced as a result of this assignment. `Self-critique ------------------------------------------------ Self-critique Rating: #*&!