Question

Please help immediately. ​

Answer 1

Answer:

Hey mate! here is your answer.

Step-by-step explanation:

Number of Red balls = 4

Number of Blue balls = 6

Number of Pink balls = 8

Total number of balls = 4 + 6 + 8 = 18

Required probability

=4/18×11/20+6/18×11/20=11/20[4/18+6/18]=11/20×10/18=11/36.So your correct option is E .

Hope it's helpful.

Please mark me as brainlist.

Answer 2

QUESTION :–

▪︎ A bag contains 4 red balls, 6 blue balls and 8 pink balls. 1 ball is drawn at random and replace with 3 pink balls. What is the probability that the first ball drawn was either red or blue in colour and the second ball drawn was pink in colour.

A) 12/21

B) 13/17

C) 11/30

D) None of these

ANSWER :–

GIVEN :–

• A bag contains 4 red balls, 6 blue balls and 8 pink balls.

• 1 ball is drawn at random and replace with 3 pink balls.

TO FIND :–

• What is the probability that the first ball drawn was either red or blue in colour and the second ball drawn was pink in colour .

SOLUTION :–

• n(red balls) = n(r) = 4

• n(blue balls) = n(b) = 6

• n(pink balls) = n(p) = 8

• Total balls = 18

▪︎ So that –

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$\:\: \implies \bf \: Probability = p(first \:\:ball \: \: was \: \: red \: \: and \: \: second \: \: was \: \: pink) + p(first \:\:ball \: \: was \: \: blue \: \: and \: \: second \: \: was \: \: pink)$

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$\:\: \implies\bf \: Probability = \dfrac{ ^{4} c_{1}}{^{18} c_{1}} \times\dfrac{ ^{11} c_{1}}{^{20} c_{1}} + \dfrac{ ^{6} c_{1}}{^{18} c_{1}} \times\dfrac{ ^{11} c_{1}}{^{20} c_{1}}$

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$\:\: \implies\bf \: Probability = \dfrac{4}{18} \times\dfrac{11}{20} + \dfrac{6}{18} \times\dfrac{11}{20}$

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$\:\: \implies\bf \: Probability = \dfrac{44}{360}+\dfrac{66}{360}$

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$\:\: \implies\bf \: Probability = \dfrac{44 + 66}{360}$

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$\:\: \implies\bf \: Probability = \dfrac{110}{360}$

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$\:\: \implies\bf \: Probability = \dfrac{11}{36}$

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$\:\: \implies \large{ \boxed{\bf Probability = \dfrac{11}{36}}}$

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▪︎ Hence , Option (D) is correct.