since 1/10 + 2/3 + 1/7 < 1 , there have to be other colours, and yes, it is possible
1/10 + 2/3 + 1/7 = 191/210
minimum counters is 210
191/210 = x/1000
210x = 191000
x = 909.5
but we have to round down since we can't have partial counter, so
909 counters
A box contains different coloured counters, with P(purple)= 10%, P(yellow)= 2/3 and P(orange) 1/7.
a. Is it possible to obtain a colour other than purple, yellow or orange?
b. What is the minimum number of counters in the box
c. If the box cannot fit more than 1000 counters, what is the maximum number of counters in the box?
1 answer
2 answers