The letters A and B are both independent of each other. The chances of a letter

Question

The letters A and B are both independent of each other. The chances of a letter being chosen can be shown in the expression A=3B/2. Suppose there is a sequence of 50 letters.

1. How many possible sequences of letters are there?

2. What is the probability that a randomly chosen sequence has:

i) all A's in the sequence (A-A-A-A-A-...)

ii) Alternating A's and B's in the sequence (A-B-A-B-A-B-...)

iii) Alternating sequences of 3 A's and 2 B's (A-A-A-B-B-A-A-A-B-B-...)

iv) All B's in the sequence (B-B-B-B-B-...)

3. What is the probability that a randomly chosen sequence has:

i) 50 A's and 0 B's

ii) 25 A's and 25 B's

iii) 30 A's and 20 B's

iv) 0 A's and 50 B's

Explanation / Answer

here P(A)=3P(B)/2

and P(A)+P(B) =1

3*P(B)/2+P(B) =1

P(B) =2/5

P(A) =3/5

i) P(all A's in sequence) =(3/5)50 =8.08*10-12

(ii) P(alienating A's and B's) =P(25 A's and 25 B's) =(3/5)25*(2/5)25 =3.2*10-16

(iii)P(Alternating sequences of 3 A's and 2 B's ) =P(30 A's and 20 B's) =(3/5)30*(2/5)20 =2.43*10-15

(iv) All B's in the sequence =(2/5)50 =1.27*10-20

3)  probability that a randomly chosen sequence has

(i) 50 A's and 0 B's =(50!/(50!*0!))*(3/5)50*(2/5)0 =(3/5)50 =8.08*10-12

(ii) 25 A's and 25 B's =(50!/(25!*25!))*(3/5)25*(2/5)25 =0.04046

(iii) 30 A's and 20 B's =(50!/(30!*20!))*(3/5)30*(2/5)20 =0.01509

*iv) 0 A's and 50 B's =(50!/(0!*50!))*(3/5)0*(2/5)50 =3.2*10-16

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