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anonymous
 one year ago
A Dallasarea radio station plays songs from a specific, fixed set of artists. The station has no DJ; instead, a computer randomly selects which songs to play. The songs themselves are picked randomly, and the same song may be played many times in a row. In the set of songs, 45% are sung by a female singer, 40% are sung by a male singer, and 15% are instrumental with no vocals. What is the probability that a particular set of three songs contains one male singer, one female singer, and one instrumental? (Hint: Be aware that there are multiple ways to achieve this pattern of songs.)
anonymous
 one year ago
A Dallasarea radio station plays songs from a specific, fixed set of artists. The station has no DJ; instead, a computer randomly selects which songs to play. The songs themselves are picked randomly, and the same song may be played many times in a row. In the set of songs, 45% are sung by a female singer, 40% are sung by a male singer, and 15% are instrumental with no vocals. What is the probability that a particular set of three songs contains one male singer, one female singer, and one instrumental? (Hint: Be aware that there are multiple ways to achieve this pattern of songs.)

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ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0Let's say F,M & I represent a song by Females, Males or Instrumentals. There are many ways to 3 songs can be played: FFF MMM III FFI and so on But we don't want all those combinations, we want only one type that has only one F, one M and one I: FMI MFI IFM IMF MIF FIM We find six ways. We can also find six by counting: three ways that the 1st song can be played, that leaves 2 ways for the second song to be played and then only one way for the last (once we already know what type of songs were played for the 1st two, there is only one way for the last), for a total of 3*2*1=3!=6. The probability of F is 0.45, the probability of M is 0.40 and the probability of I is 0.15. So each of the six combinations above have the same probability \(0.45\times0.40\times0.15\) Each or these possibilities is independent so we have $$ P(FMI)+P(MFI)+P(IFM)+P(IMF)+P(MIF)+P(FIM)\\ =6\times \left(0.45\times0.40\times0.15\right ) $$ Does this make sense?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Kind of, so u just multiply 6×0.45×0.40×0.15 ?
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