A box contains 3 gold coins and 4 silver coins. Coins are drawn one by one without replacement until all the gold coins are drawn. If a/b is the probability that the number of draws required is more than 4, then b-a value is?
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Answer:
I have it for only 2 gold coins
Step-by-step explanation:
Let E
1
and E
2
be the events that the boxes I and II are chosen respectively.
⟹P(E
1
)=
2
1
=P(E
2
)
Let A be the event that the coin drawn is of gold.
⟹P(A∣E
1
)=P(a gold coin from box I)=
2
2
=1
P(A∣E
2
)=P(a gold coin from box II)=
2
1
Probability that the other coin in the box is of gold = The probability that the gold coin is drawn from box I=P(E
1
∣A)
By using Bayes' theorem,
P(E
1
∣A)=
P(E
1
)P(A∣E
1
)+P(E
2
)P(A∣E
2
)
P(E
1
)P(A∣E
1
)
=
2
1
×1+
2
1
×
2
1
2
1
×1
=
3
2
Hence, P(A∣E
1
)=
3
2
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