Under standard assumptions about the drawing process, you only need 10 numbers, not 1024: P(the urn initially contained ten white balls), P(the urn initially contained nine white balls and one red one), P(the urn initially contained eight white balls and two red ones), and so on through P(one white ball and nine red ones). (P(ten red balls) equals 1 minus everything else.)
P(RWRWWRWRWW) is then P(4R, 6W) divided by the appropriate binomial coefficient.
Under standard assumptions about the drawing process, you only need 10 numbers, not 1024: P(the urn initially contained ten white balls), P(the urn initially contained nine white balls and one red one), P(the urn initially contained eight white balls and two red ones), and so on through P(one white ball and nine red ones). (P(ten red balls) equals 1 minus everything else.) P(RWRWWRWRWW) is then P(4R, 6W) divided by the appropriate binomial coefficient.