A die is loaded in such a way that an even number is twice as likely to occur as an odd number. If is the event that a number less than 4 occurs on a single toss of the die, find . (Hint: Start by assigning weights to each one of the possible outcomes in this experiment. If p is the probability of an odd number, 2p is the probability of an even number, and the sum of probabilities of all possible outcomes (i. e., over the whole sample space, must be 1.)

4/9

Step-by-step explanation:

Given that :

Die is designed such that even numbers are twice as likely to occur as odd number.

Hence,

Sample space :

1, 2, 2, 3, 4, 4, 5, 6, 6

Total possible outcomes = 9

Probability that a number less than 4 occurs on a single toss of the die.

Probability = required outcome / Total possible outcomes

Required outcome = number less than 4 [1, 2, 2, 3]

Hence,

P(obtaining a number less than 4):

(Number of possible outcomes less than 4 / total number of events in sample space)

= 4 / 9

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5