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1. A box contain 3 red,5 black and 2 white balls. One ball is drawn randomly from the box.After putting the ball back into the box,one ball is again drawn randomly.Find the probability that:
a) the two balls drawn are both red;
b)the two balls drawn are of the same colours;
c) the two balls drawn are of different colour.
2. Consider the following game in which a player selects (or draws) a card from a deck of card (52 cards),If a spade is selected,the player win $2 . If a Queen or King is selected, the player win $8.(If a spade Queen or King is selected,the player also win $8.)However,if the player selects other card,it costs the player $4.If the player plays 10 times of the game,how much will he be expect to gain (or loss)?
Maths
2010-01-07 4:08 am
回答 (2)
2010-01-07 4:44 am
✔ 最佳答案
1a)P(the two balls drawn are both red)= [3/(3+5+2)]^2 = 0.09
1b)P(the two balls drawn are of the same colours)
= P(2 red) + P(2 black) + P(2 white)
= 0.09 + (5/10)^2 + (2/10)^2
=0.38
1c)P(the two balls drawn are of different colour)
= 1 - P(the two balls drawn are of the same colours)
= 1 - 0.38
= 0.62
2) P(win $2 i.e. a spade is selected but not Q or K) = 11/52
P(win $8 i.e. a Queen or King is selected) = 8/52 = 2/13
P(loss $4 i.e. not a spade and not Q or K is selected) = 1 - 11/52 - 2/13. = 33/52
Plays 1 times of the game he is expect to gain :
2(11/52) + 8(2/13) - 4(33/52) = $ -23/26
Plays 10 times he is expected to loss $ 230/26 = $8.85 (2 dec.)
2010-01-07 4:54 am
1a) 3/10 * 3/10 = 9/100
1b) 9/100 + 5/10 * 5/10 + 2/10 * 2/10 =19/50
1c) 1 - 19/50 = 31/50
1b) 9/100 + 5/10 * 5/10 + 2/10 * 2/10 =19/50
1c) 1 - 19/50 = 31/50
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https://hk.answers.yahoo.com/question/index?qid=20100106000051KK01474