First one digit is chosen uniformly at random from f1; 2; 3; 4; 5g and is removed from the set; then a second digit is chosen uniformly at random from the remaining digits. What is the probability that an odd digit is picked the second time?

Respuesta :

Answer:

[tex]\frac{3}{5}[/tex]

Step-by-step explanation:

Probability of choosing an odd number in the second turn is the sum of probabilities of choosing an odd number in second turn given that an odd number or an even number is picked in first turn.

Probability of getting an odd number in the first turn out of 1,2,3,4,5 is [tex]\frac{3}{5}[/tex]

Probability of getting an even number in the first turn out of 1,2,3,4,5 is [tex]\frac{2}{5}[/tex]

Probability of getting an odd number in second turn given that an odd number was picked in the first turn (remaining : 2 odd numbers out of 4) is [tex]\frac{1}{2}[/tex]

Probability of getting an odd number in second turn given that an even number was picked in the first turn (remaining : 3 odd numbers out of 4) is [tex]\frac{3}{4}[/tex]

Total probability is [tex]\frac{3}{5} \times \frac{1}{2}  +  \frac{2}{5} \times \frac{3}{4}  =  \frac{3}{5}[/tex]