2 b only 2 strings have the maximum number of

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2. (b) Only 2 strings have the maximum number of transitions (i.e., n - 1 ). If we fix the first bit, then the rest of the string is fixed since every two consecutive bits must be different. Now since the first bit can take only two values ( 0 or 1 ), we deduce that the number of strings having the maximum number of transitions is 2. For example if n = 4 , the maximum number of transitions is 3 and the two strings having 3 transitions are 0101 or 1010 . 1
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2. (c) The number of binary strings with length n with exactly k ( k n - 1 ) transitions is 2 ( n - 1 k ) . Indeed think of the transitions as separators to be placed between the bits, since there are n bits there are n - 1 possible spaces between these bits, hence there are ( n - 1 k ) possible configurations for placing the k transitions. Once the positions of the transitions have been fixed, the binary string will become completely specified by choosing the first bit. Since there are two possible choices for the first bit (0 or 1), we deduce that the total number of binary strings with length n with exactly k must be 2 ( n - 1 k ) . 2. (d) The number of binary strings with length n that begin with 1 and have exactly k transitions is ( n - 1 k ) . Indeed, note that by the symmetry of the problem, the number of binary strings with length n with exactly k transitions that begin with 0 (say n 0 ( k ) ) is equal to the number of binary strings with length n with exactly k transitions that begin with 1 (say n 1 ( k ) ), i.e., n 1 ( k ) = n 0 ( k ) . Because any string must either start with a 0 or with a 1, we deduce from part (c) that 2 ( n - 1 k ) = n 0 ( k ) + n 1 ( k ) . In other words n 1 ( k ) = ( n - 1 k ) . 2. (e) The total number of binary strings of length n is 2 n . Indeed for the first bit we have two choices (0 or 1). Once the first bit is fixed we have two possible choices for the second bit (0 or 1), so that at each stage there are two choices, hence from the counting principle, the total number of strings will be 2 × ... × 2 | {z } n times = 2 n . 3. (a) For 0 x 1 , we have F X ( x ) = Pr( X x ) = Pr( dart falls within a circle of radius x ) = Area of the circle with radius x Area of unit circle = πx 2 π = x 2 .
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