a bag contains 5 red and 3 green balls. another bag contains 4 red and 6 green balls. if one ball is drawn from each bag. find the probability that one ball is red and one is green.
Answers
Answered by
0
Answer:
Total no. Of balls in both d bags= 5+3+4+6 =18
Total no. Of Red balls=5+4=9
Total no. Of green balls=3+6=9
Probability= fav outcomes/total outcomes
P(red ball)=9/18=1/2
P(green ball)=9/18=1/2
Hpe it hlps u
Answered by
0
Answer:
21 / 40
Step-by-step explanation:
Use:
- P( a AND b ) = P(a) × P(b) [ for independent events a and b ]
- P( a OR b ) = P(a) + P(b) [ for disjoint events a and b ]
P( one R and one G )
= P( (R from bag 1 AND G from bag 2) OR (G from bag 1 AND R from bag 2) )
= P(R from bag 1 AND G from bag 2) + P(G from bag 1 AND R from bag 2)
= P(R from bag 1) × P(G from bag 2) + P(G from bag 1) × P (R from bag 2)
= 5/8 × 6/10 + 3/8 × 4/10
= 5/8 × 3/5 + 3/8 × 2/5
= 15/40 + 6/40
= 21 / 40
Similar questions