An urn contains 5 red and 7 black balls. Two balls are drawn at random without replacement. Find the probability of getting (b) two red balls. (ii) two black balls. (iv) first red and second black ball. (ii) one red and one black ball.
Answers
Step-by-step explanation:
ANSWER
ANSWEROn each drawing, total balls =11
ANSWEROn each drawing, total balls =11Red balls =4,
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) =
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 11
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 ×
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 11
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 =
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 121
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)=
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 11
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117 ×
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117 × 11
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117 × 117
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117 × 117
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117 × 117 =
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117 × 117 = 121
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117 × 117 = 12149
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117 × 117 = 12149
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117 × 117 = 12149
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117 × 117 = 12149 (iii) red ball can be in first or second draw
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117 × 117 = 12149 (iii) red ball can be in first or second draw⇒ P(one red and one black ball) = 2×(
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117 × 117 = 12149 (iii) red ball can be in first or second draw⇒ P(one red and one black ball) = 2×( 11 )(
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117 × 117 = 12149 (iii) red ball can be in first or second draw⇒ P(one red and one black ball) = 2×( 11 )( 11
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117 × 117 = 12149 (iii) red ball can be in first or second draw⇒ P(one red and one black ball) = 2×( 11 )( 117
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117 × 117 = 12149 (iii) red ball can be in first or second draw⇒ P(one red and one black ball) = 2×( 11 )( 117
ANSWEROn each drawing, total balls =11Red balls =4,Black balls =7(i) P(two red balls) = 114 × 114 = 12116 (ii) P(two black balls)= 117 × 117 = 12149 (iii) red ball can be in first or second draw⇒ P(one red and one black ball) = 2×( 11 )( 117
Given: An urn contains 5 red and 7 black balls. Two balls are drawn at random without replacement.
To find: Find the probability of getting (i) two red balls (ii) two black balls (iii) first red and second black ball and (iv) one red and one black ball.
Solution:
The urn has 5 red balls and 7 black balls which means the total number of balls is 12.
The possibility of choosing 2 red balls from 5= 5c2
The possibility of choosing 2 black red balls from 7= 7c2
The probability of getting 2 red balls= 5c2/12c2= 10/66= 5/33.
The probability of getting 2 black balls= 7c2/12c2= 21/66= 7/22
The probability of getting first red and second black ball = 5c1 into 7c1/12c2= 35/66
The probability of getting one red and one black ball= 5c1 into 7c1/12c2 into 2= 35/66 into 2= 35/33
The probability of getting (i) two red balls= 5/33
(ii) two black balls = 7/22
(iii) first red and second black ball= 35/66
(iv) one red and one black ball= 35/33