To start think about \(a/b=34.06\) as given OR fixed data. Then, we have many possibilities to express this as a fraction, namely, \(3406/1000\) OR \(1703/50\) OR \((1703*3)/(50*3)\). In general, we are interested only in one out of two options for the remainder with the given fraction, and this is when we multiply both numerator and denominator by the even number. Hence, out of two options of coefficients (multipliers) of both numerator and denominator, there can be odd and even multipliers possible for the resultant number \(34.06\). P(remainder=6) is \(1/2\), because only the even option out two options (odd and even) results in the remainder of 6. Answer is
CAhasunhabib999 wrote:
If a and b are both integers \(\frac{a}{b}=34.06\) and what is the probability that the remainder of \(\frac{a}{b}\) is a multiple of 6?
a. 0.3
b. 0.45
c. 0.5
d. 1
e. 0.66