Answer: (a) 0.000028 (b) 0.001479
Explanation:
Given : Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.
Number of ways to form a licence plate having three letters followed by three digits :-
![(26)^3*(10)^3=17576000](https://img.qammunity.org/2020/formulas/mathematics/college/rageizldmerxaekc6o6gm0xbvyfr1h879j.png)
Number of ways to form a licence plate having two letters followed by four digits :-
![(26)^2*(10)^4=6760000](https://img.qammunity.org/2020/formulas/mathematics/college/tcd24374afy12ciiyrlt0a6t2hegw04cy7.png)
Total number of ways to form a licence plate : 17576000+6760000=24336000
(a) If a license place consists of 7777, then it must belongs to "two letters followed by four digits."
Then, the number of ways to make a license plate having number 7777 will be :_
![26*26*1^2=676](https://img.qammunity.org/2020/formulas/mathematics/college/lvmj7ppvt7hkuyt7h92vix0gy4yhrcyzbl.png)
Now, the probability that a randomly chosen plate contains the number 7777 :_
![=(676)/(24336000)=0.00002777777\approx0.000028](https://img.qammunity.org/2020/formulas/mathematics/college/p1j0dc7xs9ygoq8az4zvzghtaky827n6tu.png)
(b) Consider ME as 1 thing.
The number of ways to make a license plate contains the substring ME will be :_
![(26)*1* (10)^3+(1)*(10)^4=26000+10000=36000](https://img.qammunity.org/2020/formulas/mathematics/college/xe7tujb6ueiukjo6bsn6002yimjnn84iq0.png)
Now, the probability that a randomly chosen plate contains the substring ME :-
![=(36000)/(24336000)=0.001479](https://img.qammunity.org/2020/formulas/mathematics/college/m72kmsq9e0d0hdy0b5zctnh0skb7nnxfs7.png)