if 6 coins are tossed simultaneously what is the probability of getting atleast four heads

• n(S)= 2^6= 64
  n(E)= 6C4 + 6C5 + 6C6 = 15+6+1= 22
  p(E)= n(E)/n(S)= 22/64= 11/32(Ans)
• 10 years agoHelpfull: Yes(46) No(7)
• generally the formula is 2^n where n is the number of coins to be tossed. here n=6
  2^6=64
  then start from {hhhhhh,hhhhht,hhhhtt,....etc}from this have to calculate the 4,5,6 heads
• 10 years agoHelpfull: Yes(4) No(7)
• the outcomes will be like
  hhhhtt-1/64 [for 1 head probability is 1/2 so for hhhhtt the probability wud be 1/2*1/2*1/2*1/2*1/2*1/2
  similarly possible outcomes hhtthh,tthhhhh,hthhth,hhhtht,hthhhth.............hththh upto 22 times
  =22*1/64=22/64
  =11/32 (Ans)
• 10 years agoHelpfull: Yes(0) No(2)
• 6C4= 6*5*4*3/4*3*2*1=15
• 10 years agoHelpfull: Yes(0) No(0)
• (6c4*6c2+6c5*6c1+6c6)/6c4
• 9 years agoHelpfull: Yes(0) No(0)
• n(s) = 2^6 = 64
  then 4 head + 5 head + 6 head
  6C4 + 6C5 +6C6 = 22
  THEN
  22/64 = 11/32
  34.375
• 3 Months agoHelpfull: Yes(0) No(0)