When we toss a penny, experience shows that the probability (long term proportion) of a head is close to 1-in-2. suppose now that we toss the penny repeatedly until we get a head. what is the probability that the first head comes up in an odd number of tosses (one, three, five, and so on)?

2/3 or 0.66666
       
This is a sum of an infinite series problem. A sequence of 1 will happen with a probability of 0.5 A sequence of 3 will happen with a probability of 1/2^3, 1/8, = 0.125 In general we have an infinite series of 1/2^1 + 1/2^3 + 1/2^5 + ... + 1/2^(2n-1) where n >= 1 The sum of such a series with a constant ratio between sequential terms is S = s1/(1-r) where s1 = first term in the series r = ratio between terms. The value for s1 = 0.5 as shown above and the 2nd term is 0.125. So r = 0.125 / 0.5 = 0.25 And the sum of the infinite series is S = s1/(1-r) S = 0.5/(1 - 0.25) S = 0.5/0.75 S = 2/3 S = 0.666..66 So the probability of the first head coming up in an odd number of tosses is 2/3, or 66.6%

A) for the n row: 22+(n-1)×6therefore,b) 100-22=78that means (n-1)×6=7878÷6=1313+1=1414th row has 100 seats

What are the choices