Question

A box contains 15 resistors. Ten of them are labeled 50 Ω and the other five are labeled 100 Ω.

Answer 1

Answer:

a) 0.3333

b) 0.2381

c) 0.0952

d) 0.4286

e) 0.4523

f) 0.2637

Step-by-step explanation:

The concept applied here is dependent probability or probability without REPLACEMENT

• Given 15 resistors = Total

• 50 Ω  = 10 resistors

• 100 Ω = 5 resistors

a) What is the probability that the first resistor is 100Ω?

P(first resistor 100Ω) = number of 100Ω / Total number of resistors

= 5/15 = 0.3333

b) P( second resistor 100 Ω, the first resistor 50 Ω?

= Since second resistor of 50Ω was drawn first, the total number of resistors would have reduced, but the 100Ω resistor still maintains its number

= 10/15 x 5/14 = 0.2381

c) P( second resistor is 100 Ω, the first resistor is 100 Ω)

In this case, since first resistor is drawn, the number of 100 Ω would have reduced so also the total number of resistors

= 5/15 x 4/14 = 0.0952

d) P( that the first two resistors are both 50Ω) = Implies picking 2- 50Ω

= 10/15 x 9/14 = 0.4286

e) P(total of two resistors are selected from the box) = P( first 100Ω and second- 50Ω) or P( First 50Ω and second 100Ω) = 10/15 x 5/14 + 5/15 x 9/14 = 0.4523

f) P( more than three resistors are selected from the box) = 1 - P( Less than or equals to three resistors) - it implies that the consideration is for the 100Ω

= P( 1resistor) - P( 2- Resistor) - P( 3- Resistor)

1 - 5/15 - (10/15 + 5/14) - (10/15 + 9/14 + 5/13)

= 0.2637