Probability

Z

Zelda22

Junior Member
Joined
Mar 30, 2022
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136
Drawing without replacement
BBGGG

1-EVENT: G, then B = 3/10

G , B
3/5 * 2/4 = 6/20 = 3/10

2-EVENT: Second is G =3/10

If B, then G
2/5 * 3/4= 6/20= 3/10

Should I add 3/10 + 3/10 ??? or should be just 3/10 ?

if G, then G
3/5 * 2/4=6/20=3/10

3-EVENT: Both same = 2/5 ???

If B, then B
2/5 *1/4= 2/20=1/10

If G, then G
3/5 * 2/4=6/20=3/10

What should I do now?
add?
1/10 + 3/10= 4/10=2/5 ???

Please help. thanks
 
Last edited:
Zelda22 said:
Drawing without replacement
BBGGG

1-EVENT: G, then B = 3/10

G , B
3/5 * 2/4 = 6/20 = 3/10

2-EVENT: Second is G =3/10

If B, then G
2/5 * 3/4= 6/20= 3/10

Should I add 3/10 + 3/10 ??? or should be just 3/10 ?

if G, then G
3/5 * 2/4=6/20=3/10

3-EVENT: Both same = 2/5 ???

If B, then B
2/5 *1/4= 2/20=1/10

If G, then G
3/5 * 2/4=6/20=3/10

What should I do now?
add?
1/10 + 3/10= 4/10=2/5 ???

Please help. thanks
Click to expand...
It's hard to follow what you're being asked to find. Can you post a picture of the original problem?
 
Zelda22 said:
Here it is, thank you for helping
Click to expand...
1649798822432.png
Now, I understand. Thanks for the picture.

Yes, you need to add the two probabilities together because for event C. How many ways can we get Green as a second draw? We can have Green & Green or Blue and Green. So
P(C)=P(draw green 2nd) = P(green 1st and green 2nd) + P(blue 1st and green 2nd)

Can you use similar logic for event D?
 
If there is a 'bag' containing markers[imath]BBGGG [/imath] and two are drawn without replacement
then the probability they are the same colour is [imath]\mathcal{P}(B_1B_2\cup G_1G_2)=\dfrac{2}{5}\cdot\dfrac{1}{4}+\dfrac{3}{5}\cdot\dfrac{2}{4} [/imath]
 
For event C
Second G

I'm not sure if I'm doing it correctly?

If the first is B, then G
2/5 * 3/4 = 6/20= 3/10

If the first is G, then G
3/5 * 2/4 = 6/20= 3/10

add 3/10 + 3/10 = 6/10= 3/5 ???
 
Last edited:
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