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
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
