Question

The box contains 14 green and 19 black pencils. pencils are picked up randomly one by one without replacement. what is the probability that 2 black and 3 green pencils are picked up in first four pick ups ?

Answer 1

Answer on Question #48506-Math-Statistics and Probability

The box contains 14 green and 19 black pencils. Pencils are picked up randomly one by one without replacement. What is the probability that 2 black and 3 green pencils are picked up in first five pickups?

Solution

$$\begin{array}{l} P(2 \text{ black and 3 green pencils}) = \\ = P(BBGGG) + P(BGBGG) + P(BGGBG) + P(BGGBB) + P(GBBGG) + \\ + P(GBGBG) + + P(GBGGB) + P(GGBBG) + P(GGBGB) + P(GGGBB) = \\ = \frac{19}{33} \cdot \frac{18}{32} \cdot \frac{14}{31} \cdot \frac{13}{30} \cdot \frac{12}{29} + \frac{19}{33} \cdot \frac{14}{32} \cdot \frac{18}{31} \cdot \frac{13}{30} \cdot \frac{12}{29} + \frac{19}{33} \cdot \frac{14}{32} \cdot \frac{13}{31} \cdot \frac{18}{30} \cdot \frac{12}{29} + \frac{19}{33} \cdot \frac{14}{32} \cdot \frac{13}{31} \cdot \frac{12}{30} \cdot \frac{18}{29} + \frac{14}{33} \\ \cdot \frac{19}{32} \cdot \frac{18}{31} \cdot \frac{13}{30} \cdot \frac{12}{29} + \frac{14}{33} \cdot \frac{19}{32} \cdot \frac{13}{31} \cdot \frac{18}{30} \cdot \frac{12}{29} + \frac{14}{33} \cdot \frac{19}{32} \cdot \frac{13}{31} \cdot \frac{12}{30} \cdot \frac{18}{29} + \frac{14}{33} \cdot \frac{13}{32} \cdot \frac{19}{31} \cdot \frac{18}{30} \cdot \frac{12}{29} \\ + \frac{14}{33} \cdot \frac{13}{32} \cdot \frac{19}{31} \cdot \frac{12}{30} \cdot \frac{18}{29} + \frac{14}{33} \cdot \frac{13}{32} \cdot \frac{12}{31} \cdot \frac{19}{30} \cdot \frac{18}{29} = \frac{10 \cdot 19 \cdot 18 \cdot 14 \cdot 13 \cdot 12}{33 \cdot 32 \cdot 31 \cdot 30 \cdot 29} = 0.262 \\ \end{array}$$

Answer: 0.262.