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The bag contains,

Red (R) marbles is 9, Green (G) marbles is 7 and Blue (B) marbles is 4,

Total marbles (possible outcome) is,


\text{Total marbles = (R) + (G) +(B) = 9 + 7 + 4 = 20 marbles}

Let P(R) represent the probablity of picking a red marble,

P(G) represent the probability of picking a green marble and,

P(B) represent the probability of picking a blue marble.

Probability , P, is,


\text{Prob, P =}\frac{required\text{ outcome}}{possible\text{ outcome}}
\begin{gathered} P(R)=(9)/(20) \\ P(G)=(7)/(20) \\ P(B)=(4)/(20) \end{gathered}

Probablity of drawing a Red marble (R) and then a blue marble (B) without being replaced,

That means once a marble is drawn, the total marbles (possible outcome) reduces as well,


\begin{gathered} \text{Prob of a red marble P(R) =}(9)/(20) \\ \text{Prob of }a\text{ blue marble =}(4)/(19) \\ \text{After a marble is selected without replacement, marbles left is 19} \\ \text{Prob of red marble + prob of blue marble = P(R) + P(B) = }(9)/(20)+(4)/(19)=(251)/(380) \\ \text{Hence, the probability is }(251)/(380) \end{gathered}

Hence, the best option is G.

User Ruud De Jong
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