Probability

[Guest]

Nomusa has 30 sweets. She has 18 fruit sweets 7 aniseed sweets 5 mint sweets Nomusa is going to take at random two sweets. Work out the probability that the two sweets will not be the same type of sweet. please show all your working.

[Guest]

17/29 + 6/29 + 4/29
= 27/30
1 - 27/29 = 2/29 = 0.069

[Guest]

correction
17/29 + 6/29 + 4/29
= 27/29
1 - 27/29 = 2/29 = 0.069

[leesajohnson]

Number of fruit sweets, aniseed sweets and mint sweets:

17/29 + 6/29 + 4/29

= 27/29

1 - 27/29 = 2/29 = 0.069

[Guest]

Talking about probability: can someone explain this to me in human language? Why 14.8%?
,,THe RTP of the betting system is calculated as the ratio of net wins to net losses. Net wins are calculated as 14.8% * (\$648 - \$100). Net losses are calculated as \$100 * (100 % - 14.8 %). Therefore ((\$648 - \$100) * 14.8 %) / (\$100 * (100 % - 14,8 %)) = 95.19 %.'

[Guest]

The patterns of acceptable types:
1. Fruit and aniseed
[tex]\frac{18}{30}*\frac{7}{29}=\frac{126}{870}[/tex]

2. Aniseed and mint
[tex]\frac{7}{30}*\frac{5}{29}=\frac{35}{870}[/tex]

3. Mint and Fruit
[tex]\frac{5}{30}*\frac{18}{29}=\frac{90}{870}[/tex]

Total is
[tex]\frac{126}{870}+\frac{35}{870}+\frac{90}{870}=\frac{251}{870}[/tex]

[Guest]

[tex]P(A _{1 } \cap A _{2 })=P(A_{1})\cdot P(A_{2}|A_{1})= 18/30\cdot17/29=306/870[/tex]
[tex]P(A _{1 } \cap B _{2 })=P(A_{1})\cdot P(B_{2}|A_{1})= 18/30\cdot7/29=126/870[/tex]
[tex]P(A _{1 } \cap C _{2 })=P(A_{1})\cdot P(C_{2}|A_{1})= 18/30\cdot5/29=90/870[/tex]
[tex]P(B _{1 } \cap A _{2 })=P(B_{1})\cdot P(A_{2}|B_{1})= 7/30\cdot18/29=126/870[/tex]
[tex]P(B _{1 } \cap B _{2 })=P(B_{1})\cdot P(B_{2}|B_{1})= 7/30\cdot6/29=42/870[/tex]
[tex]P(B _{1 } \cap C _{2 })=P(B_{1})\cdot P(C_{2}|B_{1})= 7/30\cdot5/29=35/870[/tex]
[tex]P(C _{1 } \cap A _{2 })=P(C_{1})\cdot P(A_{2}|C_{1})= 5/30\cdot18/29=90/870[/tex]
[tex]P(C _{1 } \cap B _{2 })=P(C_{1})\cdot P(B_{2}|C_{1})= 5/30\cdot7/29=35/870[/tex]
[tex]P(C _{1 } \cap C _{2 })=P(C_{1})\cdot P(C_{2}|C_{1})= 5/30\cdot4/29=20/870[/tex]
[tex]S=(A _{1 } \cap A _{2 }) \cup (A _{1 } \cap B _{2 })\cup (A _{1 } \cap C _{2 })\cup (B _{1 } \cap A _{2 })\cup (B_{1 } \cap B _{2 })\cup (B _{1 } \cap C _{2 })\cup (C _{1 } \cap A _{2 })\cup (C _{1 } \cap B _{2 })\cup (C _{1 } \cap C _{2 })[/tex]
[tex]P(S)=P((A _{1 } \cap A _{2 }) \cup (A _{1 } \cap B _{2 })\cup (A _{1 } \cap C _{2 })\cup (B _{1 } \cap A _{2 })\cup (B_{1 } \cap B _{2 })\cup (B _{1 } \cap C _{2 })\cup (C _{1 } \cap A _{2 })\cup (C _{1 } \cap B _{2 })\cup (C _{1 } \cap C _{2 }))=P(A _{1 } \cap A _{2 })+P(A _{1 } \cap B _{2 })+P(A _{1 } \cap C _{2 })+P(B _{1 } \cap A _{2 })+P(B _{1 } \cap B _{2 })+P(B _{1 } \cap C _{2 })+P(C _{1 } \cap A _{2 })+P(C _{1 } \cap B _{2 })+P(C _{1 } \cap C _{2 })=306/870+126/870+90/870+126/870+42/870+35/870+90/870+35/870+20/870=870/870=1[/tex]
[tex]E= (A _{1 } \cap A _{2 })\cup (B _{1 } \cap B _{2 })\cup (C _{1 } \cap C _{2 })[/tex]
[tex]P(E)= P((A _{1 } \cap A _{2 })\cup (B _{1 } \cap B _{2 })\cup (C _{1 } \cap C _{2 }))=P(A _{1 } \cap A _{2 })+P(B _{1 } \cap B _{2 })+P( C _{1 } \cap C _{2 })=306/870+42/870+20/870=368/870=184/435[/tex]
[tex]F=E'[/tex]
[tex]P(F)=P(E')=P(S)-P(E)=1-184/435=435/435-184/435=251/435[/tex]