There are 5 fifty-cent coins, 3 one-dollar coins and 4 two-dollar coins in a purse. Now two coins are drawn from the purse.
a) Find the probability that the total value of two coins is greater than $2.6.
b) Given that the total value of two coins is greater than $2.6, find the probability that the total value of the two coins is greater than $3.1.
Probability
2011-04-22 1:03 am
回答 (2)
2011-04-22 6:20 am
✔ 最佳答案
a) P(two coins >$2.6) = P (one-dollar ,two-dollar ) + P(two-dollar ,one-dollar) + P(two-dollar ,two-dollar )= (3/12)(4/11) +(4/12)(3/11) + (4/12)(3/11)= 3/11 b)P(two coins >$3.1)= P(two-dollar ,two-dollar )= (4/12)(3/11)= 1/11Given that the total value of two coins is greater than $2.6, find the probability that the total value of the two coins is greater than $3.1.= P (two coins >$3.1)/P(two coins >$2.6)= (1/11) / ( 3/11)= 1/3
2011-04-22 1:44 am
total no. of coins = 5+3+4 = 12
$2.6 = ($1+$2) or ($2+$1) or ($2+$2)
a)
P($1+$2) = (3/12)*(4/11) = 1/11
P($2+$2) = (4/12)*(3/11) = 1/11
P(>$2.6) = 1/11 + 1/11 = 2/11
b) P($1+$2) = P($2+$2) = 1/11
given that value>2.6,
P($2+$2) = 1/2
∴P(>$3.1) = 1/2
$2.6 = ($1+$2) or ($2+$1) or ($2+$2)
a)
P($1+$2) = (3/12)*(4/11) = 1/11
P($2+$2) = (4/12)*(3/11) = 1/11
P(>$2.6) = 1/11 + 1/11 = 2/11
b) P($1+$2) = P($2+$2) = 1/11
given that value>2.6,
P($2+$2) = 1/2
∴P(>$3.1) = 1/2
參考: me
收錄日期: 2021-04-13 17:56:33
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