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A jar of 150 jelly beans contains 22 red jelly beans, 38 yellow, 20 green, 28 purple, 26 blue, and the rest are orange.Let B = the event of getting a blue jelly beanLet G = the event of getting a green jelly bean.Let O = the event of getting an orange jelly bean.Let P = the event of getting a purple jelly bean.Let R = the event of getting a red jelly bean.Let Y = the event of getting a yellow jelly bean.What is the probability of drawing a red jelly bean in the first draw WITHOUT replacing the red jelly bean, then choosing a blue jelly bean in the second draw?Use 4 decimal places for this problem. Do not round your answers for intermediate steps, only for the final answer.

Sagot :

MrRoyal

Answer:

[tex]P(R\ and\ B) = 0.0256[/tex]

Step-by-step explanation:

Given

[tex]Red = 22[/tex]

[tex]Yellow = 38[/tex]

[tex]Green = 20[/tex]

[tex]Purple = 28[/tex]

[tex]Blue = 26[/tex]

[tex]Orange = 16[/tex]

Required

Determine the probability of red then blue jelly? i.e. P(R and B)

From the question, we understand that the red jelly bean was not replaced. This means that the number of jelly beans reduced by 1 after the picking of the red jelly bean

So, we have:

[tex]P(R\ and\ B) = P(R)\ and\ P(B)[/tex]

This is then solved further as:

[tex]P(R\ and\ B) = P(R)\ *\ P(B)[/tex]

[tex]P(R\ and\ B) = \frac{n(R)}{Total}\ *\frac{n(B)}{Total - 1}[/tex]

The probability has a denominator of Total - 1 because the number of jelly beans reduced by 1 after the picking of the red jelly bean

The equation becomes:

[tex]P(R\ and\ B) = \frac{22}{150}\ *\frac{26}{150- 1}[/tex]

[tex]P(R\ and\ B) = \frac{22}{150}\ *\frac{26}{149}[/tex]

[tex]P(R\ and\ B) = \frac{22*26}{150*149}[/tex]

[tex]P(R\ and\ B) = \frac{572}{22350}[/tex]

[tex]P(R\ and\ B) = 0.02559284116[/tex]

[tex]P(R\ and\ B) = 0.0256[/tex]

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