The probability is 1% that an electrical connector that is kept dry fails during the warranty period of a portable computer. if the connector is ever wet, the probability of a failure during the warranty period is 3%. if 90% of the connectors are kept dry and 10% are wet, what proportion of connectors fail during the warranty period? round your answer to three decimal places (e. g. 98.765). enter your answer in accordance to the question statemententer your answer in accordance to the question statement

0.012 or 1.2%

Step-by-step explanation:

Let the total number of connectors be "x".

90% of the connectors are kept dry. This means, the number of dry connector is 90% of x = 0.90x

10% of the connectors are wet. This means, the number of wet connectors is 10% of x = 0.10x

There is 1% probability that the dry connectors will fail. So, out of 0.90x, only 1% of the connectors are expected to fail.

So, number of dry connectors, expected to fail = 1% of 0.90x = 0.01 (0.90x) = 0.009x

There is 3% probability that wet connectors will fail. So, out of 0.10x, only 3% of the connectors are expected to fail.

So, number of wet connectors expected to fail = 3% of 0.10x = 0.03 (0.10x) = 0.003x

Total number of connectors that will fail = 0.009x + 0.003x = 0.012x

From the "x" connectors "0.012x" connectors will fail, which means 1.2% of the connectors will fail during the warranty period.

idk

step-by-step explanation: i just don't

step-by-step explanation: