Two balls are drawn simultaneously at random; 6 red, 4 white & 5 blues. What is P(one is white and one is red)?
回答 (6)
Total number of balls = 6 + 4 + 5 = 15
P(one is white and one is red)
= P(the first is white and the last is red) + P(the first is red and the last is white)
= (4/15) × (6/14) + (6/15) × (4/14)
= 8/35
Total possible outcomes N = 15 = n(R) + n(W) + n(B) = 6 + 4 + 5 from a single draw of two
We are looking for p(R and W) = p(W) + p(R) + p(B) - p(W and R) - p(W and B) - p(B and R) - p(W and R and B); where p(W and R and B) = 0 because we're drawing only two balls. DRAW a Venn diagram to confirm this. [See source.]
So we have p(R and W) = 1 - (4/15)*(6/15) - (4/15)*(5/15) - (5/15)*(5/15) = 0.693 ANS.
NOTE: p(W) + p(R) + p(B) = n(W)/N + n(R)/N + n(B)/N = (n(R) + n(W) + n(B))/N = N/N = 1
參考: A Venn diagram shows you have three probability spaces, one per color, plus three joint spaces of which only the p(R and W) is what you are looking for.
Here is an alternate argument:
(ways to choose 1 red ball from 6)
(ways to choose 1 white ball from 4) /
(ways to choose 2 balls from 15)
C(6,1)C(4,1)/C(15,2)
= (6*4)/(15*14/2) = 8/35
P(one is white and one is red) = P(RW)+P(WR) = 2(4/15)(6/14)=...
收錄日期: 2021-05-01 22:27:20
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