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slava [35]
3 years ago
11

A container has 3 apple juice boxes, 4 grape juice boxes and 5 orange juice boxes. Alex picks one juice box, drinks it and then

picks another juice box. What is the probability that Alex picks grape juice and then orange juice?
Mathematics
1 answer:
Eduardwww [97]3 years ago
4 0

Hey!

Grape over total: \frac{4}{12}

After he drinks 1, there will be 1 less. That means it will be over 11

Orange over total: \frac{5}{11}

Now multiply the fractions

\frac{4}{12} \times \frac{5}{11} = \frac{20}{132} \rightarrow \frac{10}{66} \rightarrow \frac{5}{33}

That means the probability is: \frac{5}{33}

Good luck and hope this helps! :)

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. Imagine a game of 3 players where exactly one player wins in the end and all players have equal chances of being the winner. T
lbvjy [14]

Answer:

5/9

Step-by-step explanation:

Number of players = 3

number of times game is repeated = 4

P( any person wins a game ) = 1/3

P ( any person does not win a game ) = 1 - 1/3 = 2/3

P ( any person wins no game in 4 attempts ) = ( 2/3 )^4 = 16/81

<em>Note : each player has equal chance of winning </em>

<u>Find the probability that there is at least one person who wins no games </u>

lets represent the probability of each player not wining a game with alphabet A

A1 = player 1 wins no game

A2 = player 2 wins no game

A3 = players 3 wins no game

Applying the inclusion-exclusion formula

<em>P( A1 ∪ A2 U A3 )</em><em> = P(A1 ) + P(A2) + P(A3) - P( A1 ∩ A2 ) - P( A2 ∩ A3 ) - P( A1            ∩ A3 )  + P( A1 ∩ A2 ∩ A3 ) </em>

where

P( A1 ∩ A2 ) = P( A1 wins all games )

P ( A1 wins all games in 4 attempts ) = ( 1/3 )^4 = 1/81

P( A1 ∩ A2 ∩ A3 ) = P ( no players wins any game in 4 attempts ) = 0

Hence

P( A1 ∪ A2 U A3 ) = 16/81 + 16/81 + 16/81 - 1/81 - 1/81 - 1/81 - 0 = 5/9

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