A die is thrown repeatedly at most
7
7 times until we get an odd number.
What is the probability that the die is thrown an even number of times in total?
shouldn’t we take 1/3*1/3 for the first case?
Did you mean \frac 12 \cdot \frac 12?
On a standard die, half of the possible outcomes are even (2, 4, 6) and half are odd (1, 3, 5). Therefore, the probability of rolling an even number = probability of rolling an odd number = \frac 36 = \frac 12
no no, so for Even numbers it’ll be 1/3 and for odd also 1/3 right?
Why should it be 1/3?
because we are taking 1 out of 3 even numbers and same goes for odd
If you’re trying to choose one even number from a pool of three even numbers , the corresponding probability of picking an even number is 1 not \frac 13. You’re guaranteed to pick an even number because there isn’t anything else to pick from.
Anyway, for this question, you’re picking an even number from \{1,2,3,4,5,6\}. There are 3 even numbers among the 6 numbers, so the probability of picking an even number is \frac 36 = \frac 12
got it, thank you!