Suppose Frieda makes four independent hops without stopping so that each outcome is equally likely. Solution 3 (Finds Numerator and Denominator Separately) Then, calculating the probabilities of each of these cases happening, we have, so the answer is. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Denote center by M, edge by E, and corner by C. We can draw a state diagram with three states: center, edge, and corner. Solution 2 (Direct Counting and Probability States) She goes back to the center in Case 1 with probability, and to the right edge with probabilityīut, don't forget complementary counting. Subcase 1: She goes back to the left edge. Now, she can go in any 4 directions, and then has 2 options from that edge. We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since, we will ignore the leading probability.įrom the left, she either goes left to another edge ( ) or back to the center ( ). What is the probability that she reaches a corner square on one of the four hops?įirst, the frog can go left with probability. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square.
For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. 15 Video Solution by MRENTHUSIASM (English & Chinese)įrieda the frog begins a sequence of hops on a grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right.14 Video Solution by TheCALT (Casework).12 Video Solution by OmegaLearn (Using Probability States).4 Solution 3 (Finds Numerator and Denominator Separately).3 Solution 2 (Direct Counting and Probability States).
0 kommentar(er)
