A bag contains 10 gold and 8 silver coins. Two successive drawings of 4 coins are made such that i) coins are replaced before the second trial ii) the coins are not replaced before the second trial. Find the probability that the first drawing will give 4 gold and the second 4 silver coins.
Answers
Given : A bag contains 10 gold and 8 silver coins.
Two successive drawings of 4 coins are made such that i) coins are replaced before the second trial ii) the coins are not replaced before the second trial.
To Find : probability that the first drawing will give 4 gold and the second 4 silver coins.
Solution:
i) coins are replaced before the second trial
Gold = 10
Silver = 8
first drawing will give 4 gold = ¹⁰C₄/¹⁸C₄
second drawing will give 4 silver = ⁸C₄/¹⁸C₄
probability that the first drawing will give 4 gold and the second 4 silver coins. = ( ¹⁰C₄/¹⁸C₄) * ( ⁸C₄/¹⁸C₄)
= ¹⁰C₄ * ⁸C₄ /(¹⁸C₄)²
ii) the coins are not replaced before the second trial.
first drawing will give 4 gold = ¹⁰C₄/¹⁸C₄
second drawing will give = ⁸C₄/¹⁴C₄
probability that the first drawing will give 4 gold and the second 4 silver coins. = ( ¹⁰C₄/¹⁸C₄) * ( ⁸C₄/¹⁴C₄)
=( ¹⁰C₄ * ⁸C₄) /(¹⁸C₄.¹⁴C₄)
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Answer:
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