Question

In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers 'true'; if it falls tails, he answers 'false'. Find the probability that he answers at least 12 questions correctly.

Answer 1

Answer:

9/13

Step-by-step explanation:

Let E1: students residing in the hostel → P(E1) = 60% = 60/100=3/5

Let E2

: day scholars → P(E2) = 40% = 40/100=2/5

30% of hostel students get an A grade →

P (E|E1) = 30% = 30/100=3/10

20% of day scholars get an A grade →

P (E|E2) = 20% = 20/100=1/5

We need to find the probability that a student who is chosen from random that has an A grade is from the hostel.

We can use Baye's theorem, according to which P(E1|A)=P(E1)(P(A|E1)P(E1)P(A|E1)+P(E2)+P(A|E2)

Using Baye's theorem, P (E1

|A) =35.31035.310+25.15 = 9/9+4=9/13

Answer 2

Solution:

Total number of times the student tosses a coin = 20

Let Probability of success(On head mark true) be p = 1/2

Probability of failure q = 1-p = 1/2

To Find the probability that he answers at least 12 questions correctly, one can apply Bernoulli's theorem of Probability:

$p(x= 12)=20C_{12} \: {(p)}^{12} \times {(q)}^{8} \\ \\or \\ \\ p(x=12) = 20C_{12} \: {\bigg( \frac{1}{2} \bigg)}^{20}$
Since p = q in this case.

Now to find the probability of at least 12 success

$p(x \geqslant 12) = 20C_{12} \: {\bigg( \frac{1}{2} \bigg)}^{20} + 20C_{11} \: {\bigg( \frac{1}{2} \bigg)}^{20} + 20C_{10}\: {\bigg( \frac{1}{2} \bigg)}^{20} \\ \\ + 20C_{9} \: {\bigg( \frac{1}{2} \bigg)}^{20} + 20C_{8} \: {\bigg( \frac{1}{2} \bigg)}^{20} + 20C_{7} \: {\bigg( \frac{1}{2} \bigg)}^{20} \\ \\ + 20C_{6} \: {\bigg( \frac{1}{2} \bigg)}^{20} + 20C_{5} \: {\bigg( \frac{1}{2} \bigg)}^{20} + 20C_{4} \: {\bigg( \frac{1}{2} \bigg)}^{20} + 20C_{3} \: {\bigg( \frac{1}{2} \bigg)}^{20} + \\ \\ + 20C_{2}\: {\bigg( \frac{1}{2} \bigg)}^{20} + 20C_{1} \: {\bigg( \frac{1}{2} \bigg)}^{20} \\ \\ = {\bigg( \frac{1}{2} \bigg)}^{20}(20C_{1} + 20C_{2} + 20C_{3} + 20C_{4} + ... + 20C_{11} + 20C_{12})$

It is a complex calculation.we can left it as it is

Hope it helps you.