Abox contains 15 resistors. ten of them are labeled 50 ω and the other five are labeled 100 ω.

what is the probability that the first resistor is 100ω?

what is the probability that the second resistor is 100 ω, given that the first resistor is 50 ω?

what is the probability that the second resistor is 100 ω, given that the first resistor is 100 ω?

refer to exercise 3. resistors are randomly selected from the box, one by one, until a 100 ω resistor is selected.

what is the probability that the first two resistors are both 50ω?

what is the probability that a total of two resistors are selected from the box?

what is the probability that more than three resistors are selected from the box?

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

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