Sam see you when you divide your answer is always smaller than what you start with
2 answers:
You might be interested in
Answer:
a) 0.5 = 50% probability that an NPN transistor will be selected.
b) 0.3333 = 33.33% probability that it came from the cabinet that contains both types
c) 66.67% probability that it comes from the cabinet that contains only NPN transistors
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
In which
P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
P(A) is the probability of A happening.
a) What is the probability that an NPN transistor will be selected?
1/3 probability that the first cabinet is chosen. This cabinet has two transistors, both of which are NPN, so 100% probability of selecting a NPN transistor.
1/3 probability that the second cabinet is chosen. This cabinet has two transistors, both of which are PNP, so 0% probability of selecting a NPN transistor.
1/3 probability that the second cabinet is chosen. This cabinet has two transistors, one of which is NPN, so 50% probability of selecting a NPN transistor.
So
0.5 = 50% probability that an NPN transistor will be selected.
b) Given that the hobbyist selects an NPN transistor, what is the probability that it came from the cabinet that contains both types?
Here we use the conditional probability formula.
Event A: NPN transistor
Event B: From the third cabinet.
50% probability that an NPN transistor will be selected, so .
1/6 probability that it is from the third cabinet and NPN, so
The desired probability is:
0.3333 = 33.33% probability that it came from the cabinet that contains both types.
c) Given that an NPN transistor is selected what is the probability that it comes from the cabinet that contains only NPN transistors?
Either it comes from the cabinet with only NPN transistors, or it comes from the cabinet with both types of transistors. The sum of the probabilities of these outcomes is 100%. So
x + 33.33 = 100
x = 66.67
66.67% probability that it comes from the cabinet that contains only NPN transistors
Answer:
We can do it with envelopes with amounts $1,$2,$4,$8,$16,$32,$64,$128,$256 and $489
Step-by-step explanation:
- Observe that, in binary system, 1023=1111111111. That is, with 10 digits we can express up to number 1023.
This give us the idea to put in each envelope an amount of money equal to the positional value of each digit in the representation of 1023. That is, we will put the bills in envelopes with amounts of money equal to $1,$2,$4,$8,$16,$32,$64,$128,$256 and $512.
However, a little modification must be done, since we do not have $1023, only $1,000. To solve this, the last envelope should have $489 instead of 512.
Observe that:
- 1+2+4+8+16+32+64+128+256+489=1000
- Since each one of the first 9 envelopes represents a position in a binary system, we can represent every natural number from zero up to 511.
- If we want to give an amount "x" which is greater than $511, we can use our $489 envelope. Then we would just need to combine the other 9 to obtain x-489 dollars. Since
, by 2) we know that this would be possible.