Wiki-Quellcode von Lösung Urne mit Kugeln befüllen (2)
Version 1.1 von Simone Schuetze am 2026/04/29 16:11
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| author | version | line-number | content |
|---|---|---|---|
| |
1.1 | 1 | **Wahrscheinlichkeiten** |
| 2 | |||
| 3 | Zwei rote Kugeln: | ||
| 4 | |||
| 5 | {{formula}}P(rr)=\frac{x}{x+4}\cdot\frac{x-1}{x+3}=\frac{x(x-1)}{(x+4)(x+3)}{{/formula}} | ||
| 6 | |||
| 7 | Zwei verschiedenfarbige Kugeln: | ||
| 8 | |||
| 9 | {{formula}}P(rb)+P(br)={{/formula}} | ||
| 10 | |||
| 11 | {{formula}}\frac{x}{x+4}\cdot\frac{4}{x+3}+\frac{4}{x+4}\cdot\frac{x}{x+3}{{/formula}} | ||
| 12 | |||
| 13 | {{formula}}=\frac{4x}{(x+4)(x+3)}+\frac{4x}{(x+4)(x+3)}=\frac{8x}{(x+4)(x+3)}{{/formula}} | ||
| 14 | |||
| 15 | **Bedingung aufstellen** | ||
| 16 | |||
| 17 | {{formula}}P(rr)=P(rb)+P(br){{/formula}} | ||
| 18 | |||
| 19 | {{formula}}\frac{x(x-1)}{(x+4)(x+3)}=\frac{8x}{(x+4)(x+3)}{{/formula}} | ||
| 20 | |||
| 21 | **Gleichung lösen** | ||
| 22 | |||
| 23 | Beide Seiten werden mit {{formula}}(x+4)(x+3){{/formula}} multipliziert: | ||
| 24 | |||
| 25 | {{formula}}x(x-1)=8x{{/formula}} | ||
| 26 | |||
| 27 | {{formula}}x^2-x=8x{{/formula}} | ||
| 28 | |||
| 29 | {{formula}}x^2-9x=0{{/formula}} | ||
| 30 | |||
| 31 | {{formula}}x(x-9)=0{{/formula}} | ||
| 32 | |||
| 33 | **Lösung** | ||
| 34 | |||
| 35 | {{formula}}x=9{{/formula}} (da {{formula}}x=0{{/formula}} nicht sinnvoll ist) | ||
| 36 | |||
| 37 | Es müssen 9 rote Kugeln in der Urne sein. |