Une urne contient $20\%$ de boules rouges, $50\%$ de boules vertes et le reste est compos\'e de boules bleues. Les boules sont indiscernables au toucher. L'exp\'erience al\'eatoire consid\'er\'ee consiste \`a tirer une boule au hasard dans l'urne. D\'eterminer la loi de probabilit\'e de cette exp\'erience.

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Grade 6

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Q. Une urne contient

20\%

de boules rouges,

50\%

de boules vertes et le reste est compos\'e de boules bleues. Les boules sont indiscernables au toucher. L'exp\'erience al\'eatoire consid\'er\'ee consiste \`a tirer une boule au hasard dans l'urne. D\'eterminer la loi de probabilit\'e de cette exp\'erience.

Identify Color Percentages: Let's identify the total percentage of each color of balls in the urn. $\newline$ We have: $\newline$ - $20\%$ red balls $\newline$ - $50\%$ green balls $\newline$ Since there are only three colors, the percentage of blue balls can be found by subtracting the sum of the percentages of red and green balls from $100\%$ . $\newline$ $100\% - (20\% + 50\%) = 100\% - 70\% = 30\%$ $\newline$ So, there are $30\%$ blue balls.
Determine Probability Distribution: Now, we can determine the probability distribution for the experiment of drawing one ball from the urn. $\newline$ The probability of drawing a red ball (R) is $20\%$ , or $0.20$ . $\newline$ The probability of drawing a green ball (G) is $50\%$ , or $0.50$ . $\newline$ The probability of drawing a blue ball (B) is $30\%$ , or $0.30$ .
Summarize Probability Distribution: We can summarize the probability distribution as follows: $\newline$ P( $R$ ) = $0.20$ $\newline$ P( $G$ ) = $0.50$ $\newline$ P( $B$ ) = $0.30$ $\newline$ This distribution must sum up to $1$ , which is the total probability for any event. $\newline$ $0.20 + 0.50 + 0.30 = 1.00$