A die is loaded in such a way that an even number is twice as likely to occur as an odd number. If E is the event that a number less than 4 occurs on a single toss of the die, find P(E).

Step-by-step explanation:

Let the probability of an even number occurring be p and that of odd be q. From the question, p = 2q.

Now, p and q are mutually exclusive events. Hence,

p + q = 1

2q + q = 1

3q = 1

q = 1/3

p = 2q = 2/3

There are 3 odd numbers and 3 even numbers on a die. The probability of each even number is p/3 = 2/9 and the probability of each odd number is q/3 = 1/9.

The event E is the event of a 3, 2 or 1 appearing. Here, we have two odd numbers and 1 even number.

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)