Probability Question?

Question

Can you help me explain this question to a friend?

A quarter is flipped and one number is drawn from a box containing the numbers 2,4,6,8,10.

What is the probability of getting a tails or the numbers 6 or 10?

Answer 1

With the coin, there are 2 possible outcomes: heads or tails.
With the box, there are 5 possible outcomes: one per number.

Therefore, there are 2 x 5 = 10 possible outcomes total.

heads and 2
heads and 4
heads and 6
heads and 8
heads and 10
tails and 2
tails and 4
tails and 6
tail and 8
tails and 10

The answer to your question really depends on the use of the word "or." If it is an inclusive "or," then 7/10 of the results meet your criteria. If it is an exclusive "or," then only 5/10 of the results meet your criteria.

Answer 2

Not the typical probability questions I've been doing...

Here's the main hindrance in doing this probability problem. Since flipping the quarter is independent of drawing the numbers, we cannot just add them.

Let's count the number of possible choices we can get.

We can get a tails along with 2, 4, 6, 8, or 10 - or... - heads along with 2, 4, 6, 8, or 10.

Total number of possible choices is 10.

Now we have to find the number of possible ways to get tails or 6 or 10.

List the possibilities first:

tails 2 ----- heads 2
tails 4 ----- heads 4
tails 6 ----- heads 6
tails 8 ----- heads 8
tails 10 ----- head 10

1/2 + 1/10 + 1/10 = 1/2 + 1/5 = 7/10

Probability = 7/10

Let's pray I did that perfectly.

Answer 3

The probablilty of getting a 6 or 10: 2/5
The probability of getting a tails: 1/2
Now, DON'T just add them up. You need to also subtract the probability of getting both, since that's counted twice if you simply add.
P(Both)= 2/5*1/2=1/5
2/5+1/2-1/5=4/10+5/10-2/10
=7/10
Therefore, the probability is 7/10.

Answer 4

Its 9/10. Let see how we got it
Let A,B,C denote events its tail,no. is 6 and no. is 10 resp.
Then,
P(A)=1/2
P(B)=1/5
P(C)=1/5
Therefore the required probability=P(AUBUC)
=P(A)+P(B)+P(C)
=1/2+1/5+1/5
=9/10
Its the right one and so simple. I have never got a mark less in any of my probability tests, so don't worry .

Answer 5

Possible Outcomes

T,2 <==
T,4 <==
T,6
T,8 <==
T,10
H,2
H,4
H,6 <==
H,8
H.10 <==

T OR 6 OR 10 = 5/10

P(T) = 1/2
P(6 OR 10) = 2/5
P(T AND 6 OR 10) = 2/10
P(T OR 6 OR 10) = P(T) + P(6 OR 10) - 2*P(T AND 6 OR 10)
= 5/10 + 4/10 - 2(2/10) = 5/10 = 1/2