Answer to Question #303188 in Statistics and Probability for Rawan saud

Question

Q−5:

a)  An urn contains 5 red balls and 6 blue balls. A ball is drawn randomly, its colour is noted, and then the ball is replaced in the urn. If thisprocess is repeated four times: 

i. What is the probability of getting two red balls and 2 blue balls?

ii. What is the probability that all 4 balls are of the same colour?

iii. What is the probability that the first red ball is drawn in the 4th trial?

b)  Number plates in an Arab country consist of 3 letters (chosen from a specified set of 20 letters from the Arabic alphabet) and 4 digits (from 0 to 9).

i. Find the probability of a plate having identical letters.

ii. Find the probability of a plate having identical digits.

iii. Find the probability of a plate having consecutive digits.

Answer 1

a)

i.

$$\{RRBB, RBBR, RBRB, BRBR, BRRB, BBRR\}$$

$$P(2B\&2R)=6(1/5)^2(1/6)^2=1/150$$

ii.

$$\{RRRR, BBRB\}$$

$$P(4B\ or\ 4R)=(1/5)^4+(1/6)^4=1/625+1/1296$$

$$=\dfrac{1921}{810000}$$

iii.

$$\{BBBR\}$$

$$P(first\ red \ in \ 4^{th})=(1/6)^3(1/5)=1/1080$$

b) Suppose that number consists of 3 letters of Arabic alphabet followed by 4 digits.

There are $20^3\cdot10^4$ possible outcomes.

i.Consider three identical letters as one unit.

There are $20(3!)\cdot10^4$ favorable cases.

The probability of a plate having identical letters is

$$\dfrac{20(3!)\cdot10^4}{20^3\cdot10^4}=0.015$$

ii.Consider four identical letters as one unit.

There are $20^3\cdot10(4!)$ favorable cases.

The probability of a plate having identical letters is

$$\dfrac{20^3\cdot10(4!)}{20^3\cdot10^4}=0.024$$

iii.

There are $20^3\cdot2(7)$ favorable cases.

The probability of a plate having identical letters is

$$\dfrac{20^3\cdot2(7)}{20^3\cdot10^4}=0.0014$$