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A bag contains $4$ white balls and $2$ black balls , another contains $3$ white balls and $5$ black balls . If one ball is drawn from each bag, determine the probability that both are white .

$ a.)\ \dfrac13 \\ b.)\ \dfrac23 \\ \color{green}{ c.)\ \dfrac14 } \\ d.)\ \dfrac34 $

I did $\dfrac12 \times \dfrac46 + \dfrac12 \times \dfrac38=\dfrac{25}{48} $

But the answer is given as option $c.)$

I look for a short and simple way .

I have studied maths upto $12$th grade.

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  • $\begingroup$ You are selecting one ball from each bag, so the probability that you select a ball from the first bag is $1$, as is the probability that you select a bag from the second bag. Moreover, when you perform two independent tasks, the probabilities multiply. You add the probabilities when two tasks are mutually exclusive. $\endgroup$
    – N. F. Taussig
    Oct 19, 2015 at 19:24

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These are two independent events so you just need to find the probability of each and then multiply.

Probability that the ball from first bag is white=$4/6$

Probability that the ball from second bag is white=$3/8$

So the answer is $4/6\times3/8=1/4.$

What you have found is the probability that pick a white ball in this experiment: We first pick a bag at random that then pick a ball at random.

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