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Answer 1 · 2018-02-04T01:21:56

The probability that at most three of them are black #=0.3125 +0.3125+0.1563=0.7813#

Given-

Number of black buttons #=5#
Number of brown buttons #=5#
Total buttons #=10#

If one button is taken at random the probability of it to be black #=5/10=0.5#

#p=0.5#
#q=1-0.5=0.5#

In five tries, we take five buttons, we expect the following results

3 black and 2 brown
Or
2 black and 3 brown
Or
1 black and 4 brown

It is like tossing five coins, with known probability, is tossed and expect 3 or 2 or 1 head.

Use the binomial formula

#P_((r))=""^nC_r.p^r.q^(n-r)#

3 black and 2 brown

#P_((r=3))=""^5C_3xx0.5^3xx0.5^(5-3)=10 xx 0.125xx0.25=0.3125#

2 black and 3 brown

#P_((r=2))=""^5C_2xx0.5^2xx0.5^(5-2)=10 xx0.25xx0.125=0.3125#

1 black and 4 brown

#P_((r=1))=""^5C_1xx0.5^1xx0.5^(5-1)=5xx0.50 xx 0625=0.1563#

The probability that at most three of them are black #=0.3125 +0.3125+0.1563=0.7813#