TRS because of SSS, SAS, SSA, ASA, and probably others.
Brainliest?
Answer:
WELP 30 mins passed
Step-by-step explanation:
The answer would be 12 just adding 3
Ok i did just that is that ok
Answer:
A. P( 2∪3∪1 )=1/2
B. P( 2∪3∪4∪5∪6 )= 13/16
Step-by-step explanation:
A. At first we need to get all the information that we can from the question, so we focus on the probabilities given P(2∪3∪5)=1/2 and P(4∪5∪6)=1/2 and we need to use P(5)=3/16.
Notice that when you have a dice, it´s impossible to get two values at the same try. Because of this and the property of probability for events:
P(2∪3∪5)=P(2) + P(3) + P(5) - P(2∩3)(=0) - P(3∩5)(=0) - P(2∩5)(=0) + P(2∩3∩5)(=0)
P(2∪3∪5)=P(2) + P(3) + P(5)
1/2=P(2) + P(3) + 3/16
P(2)+P(3)=5/16
If we do the same with P(4∪5∪6), we get:
P(4)+P(6)=5/16
Now, with this new information we can find P(1):
P(1)= 1 - P(2∪3∪4∪5∪6)
P(1)= 1 - (P(2) + P(3) + P(4) + P(5) + P(6)) (The probability of two of them happening at the same time is zero)
P(1)= 1 - (5/16 + 5/16 + 3/16)
P(1)= 1 - 13/16
P(1)= 3/16
And with it, we can find P(2∪3∪1):
P(2∪3∪1)=P(2)+P(3)+P(1)
P(2∪3∪1)=5/16+3/16
P(2∪3∪1)=1/2
B. Notice if you want to roll anything but a 1, then the probability you are looking for is "P(2∪3∪4∪5∪6)" but we found that in part A
P(2∪3∪4∪5∪6)=13/16