A binary source generates digits 1 and 0 randomly with the probability. P(1)=0 7
ID: 3159624 • Letter: A
Question
A binary source generates digits 1 and 0 randomly with the probability. P(1)=0 75.and P(0) = 0.25. What is the probability that two 1' s and three 0's will occur in a 5-digit sequence? What is the probability that at least three l's will occur in a 5-digit sequence? A Lab has machines A1, A2, and A3 for making resistors with 80%, 90% and 60% Qualified Rates. Every hour, machines A1, A2 and A3 produce 5000, 2000 and 3000 resistors, respectively. All of the resistors are mixed together at random in one bin and packed for shipment, What is the probability that the Lab ships a resistor that is Qualified? What is probability that an acceptable resistor is from A3? An angle modulated signal with carrier frequency wt power of the modulated signal, frequency deviation, phase deviation omega_c =2 pi times 10^5 rad/s is given below: Find the power of the modulated signal, frequency deviation, phase deviation, the bandwidth of the FM signal. Phi_FM (t) = 30 cos (omega_c t + 2 * sin 3000 pi t + 10 * sin 2000 pi t)Explanation / Answer
No.5. Here, the distribution which we need to apply is binomial(5,0.75). We are considering getting 1 as a success.
Now we have to find the probability of getting 2 1s and 3 0s.
So, required probability is
P(X=2)=5C2*(0.75)2(0.25)3=0.08789
Now we need to find the probabity of getting atleast 3 1s.
So required probability=P(X=3)+P(X=4)+P(X=5)=5C3*(0.75)3(0.25)2+5C4*(0.75)4(0.25)+5C5*(0.75)5=0.26367+0.3955+0.2373=0.89647
No.6.
Total number of registers=5000+2000+3000=10000
Total number of qualified registers
=5000*(80/100)+2000*(90/100)+3000*(60/100)=4000+1800+1800=7600
The probability of selecting a registor that is qualified=7600/10000=0.76
Probability that acceptable registor is from A3=P(registor is qualified|A3 is selected)P(A3 is selected)=(1/3000)*(60/100)=0.0002
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