3. A box contains 6 blue balls and 4 red balls. Two balls are drawn from the box, one after the other, with replacement. Determine the probability that one of the balls is red, and the other ball is blue. Give your answer as a decimal rounded to the nearest hundredth. (2 marks) 4/10 .40 4. The probability that Andrew's mother will serve rice with dinner is 0.78. The probability that she will serve carrots is with dinner is 0.30. The probability that she will serve neither rice nor carrots is 0.14.

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GeorgesZ515091 GeorgesZ515091 · Answer 1 · 2022-10-31T13:03:23+01:00

The given

The probability of serving rice with dinner

[tex]P(A)=0.78[/tex]

The probability of serving carrots with dinner

[tex]P(B)=0.30[/tex]

The probability of serving neither rice nor carrot is

[tex]P(A^{\prime}\cap B^{\prime})=0.14[/tex]

The ven diagram is

2)

The probability of serving both rice and carrot is

[tex]P(A\cup B)=P(A)+P(B)-P(A\cap B)[/tex]

Recall that

[tex]P\mleft(AUB\mright)=1-P(A^{\prime}\cap B^{\prime})[/tex]

Substitute the values, we get

[tex]P\mleft(AUB\mright)=1-0.14[/tex]

[tex]P\mleft(AUB\mright)=0.86[/tex]