Eight cavities in an injection-molding tool produce plastic connectors that fall

Question

Eight cavities in an injection-molding tool produce plastic connectors that fall into a common stream. A sample is chosen every several minutes. Assume that the samples are independent. (a) What is the probability that five successive samples were all produced in cavity 1 of the mold? (b) What is the probability that five successive samples were all produced in the same cavity of the model? (c) What is the probability that four out of five successive samples were produced in cavity 1 of the mold?

Explanation / Answer

1. Each sample has a 1/8 chance of being from cavity 1
Therefore for 5 in a row the probability = 1/8 * 1/8 * 1/8 * 1/8 * 1/8 = (1/8)^5 = 1 / 32768
= 0.00003 to 5 decimal places

2. There is the same probability as in part A for all the other 7 cavities
Therefore there is 8 * (1/8)^5 probability that they all come from the same unspecified cavity
Probability = 1 / 4096
Probability = 0.00024 to 5 decimal places

3. The probability of getting any one sample from cavity 1 = 1/8
Therefore the probability of getting any one sample from any other cavity = 7/8
Therefore the probability of getting 4 from cavity one followed by one from another cavity = 1/8 * 1/8 * 1/8 * 1/8 * 7/8 = 7/32768
HOWEVER there are 5 ways that you can get ANY four from cavity 1 and one from another. These 5 ways are
1111O, 111O1, 11O11, 1O111 and O1111
Each of these 5 ways also have 7/32768 probability of occurring
Therefore P(4 from cavity 1 and one from another cavity IN ANY ORDER) = 5 * 7/32768 = 35/32768
Probability = 0.00107 to 5 decimal places

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