5 distinguishable coins each with probability of head 3/5 are tossed once .What is the probability of seeing exactly 2 tails
a)(20/9)*(3/4)4 b)NONE c)20*(3/4)4 d)(40/9)*(3/5)5

• Its 5c2* (2/5)^2*(1-2/5)^3=>10*4/25*(3/5)^3=2*4/5*(3/5)^3=8/5*(3^2/3^2)*5/5*(3/5)^2 ie multiply and divide by 3^2 and 5
  Now simplifying 40/9*(3/5)^5..a bit tricky question
• 9 years agoHelpfull: Yes(16) No(0)
• let's consider the sequence of seeing is T T H H H (exactly 2 tails).
  probability of this sequence = (2/5)*(2/5)*(3/5)*(3/5)*(3/5) [probability of tail (1-5/3).
  now, T T H H H this sequence can permute in (5!)/3!*2! ways.
  therefore total probability-- [(2/5)*(2/5)*(3/5)*(3/5)*(3/5)]*[(5!)/3!*2!]
  =>216/625[option d]
• 9 years agoHelpfull: Yes(6) No(0)
• Formula is ncr p^r (1-p)^n-r;where p is the prob. Of success.5c2*(3/5)^2*(2/5)^3..i tnk so none is the correct ans
• 9 years agoHelpfull: Yes(1) No(3)
• 5C2×(2/5)^2 ×(3/5)^3=40×(3/5)^5 /9
  Bay's theorem
• 9 years agoHelpfull: Yes(0) No(0)