number of ways that the 3 reds and the 2 whites
can be arranged = 5!/(3!2!) = 10
Now place a white at the end, then the number of ways to arrange the remaining 4 is 4!/(2!2!) = 6
so prob that the last one will be white = 6/10 = 3/5
A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white ?
2 answers
thank you very much. can u help me for other quistion pleasee