Dubai [UAE], October 18 (ANI): Mumbai Indians won the toss and opted to bat first against Kings XI Punjab (KXIP) in the Indian Premier League (IPL) here on Sunday.
Mumbai Indians are in a phenomenal form as the team won all of their five previous matches and will be looking to extend their winning streak. The Rohit Sham-led side holds the second spot on the points table with 12 points, only behind Delhi Capitals.
KXIP, on the other hand, have won just two of their eight matches and are placed on the bottom of the points table with just four points. However, the team secured a brilliant eight-wicket win over Royal Challengers Bangalore in the previous match and will aim to deliver a similar performance against Mumbai Indians.
Neither of the team has made a change in their playing XI.
Kings XI Punjab Playing XI: KL Rahul(w/c), Mayank Agarwal, Chris Gayle, Nicholas Pooran, Glenn Maxwell, Deepak Hooda, Chris Jordan, Murugan Ashwin, Mohammed Shami, Ravi Bishnoi, Arshdeep Singh;
Mumbai Indians Playing XI: Rohit Sharma(c), Quinton de Kock(w), Suryakumar Yadav, Ishan Kishan, Hardik Pandya, Kieron Pollard, Krunal Pandya, Nathan Coulter-Nile, Rahul Chahar, Trent Boult, Jasprit Bumrah. (ANI)