Answer:
{GT, GH, BT, BH, RT, RH};
1 /5;
Mutually exclusive ;
Not mutually exclusive.
Step-by-step explanation:
Given :
Green cards, G = 4
Blue cards, B = 4
Red cards, R = 2
For a coin toss :
{H, T}
A card is picked, then a coin is tossed :
Sample space :
{GT, GH, BT, BH, RT, RH}
2.) probability of picking a green card, then probability of landing a head on a coin toss
P(A) = number of required outcome / Total possible outcomes
P(A) = P(green card) * P(Head)
P(A) = (4 / 10) * (1/2)
P(A) = 4 /20
P(A) = 1/5
C.)
Both A and B are mutually exclusive, since both event A and event B cannot occur together, since both red and green cannot be picked during a single pick, this either a red is picked or green is picked, then they are A and B are mutually exclusive.
D.) Event A and C are not mutually exclusive, picking a green card, event A and picking a red or blue card, event B. Both event can happen simultaneously, hence, event A and B are not mutually exclusive.
- its the most reduced form. We cant divide 7 and 20 by the same number to not obtain number with coma. (because 7 is prime number so can be divided only by 1 and by 7)