22 votes 22 votes Four fair coins are tossed simultaneously. The probability that at least one head and one tail turn up is $\frac{1}{16}$ $\frac{1}{8}$ $\frac{7}{8}$ $\frac{15}{16}$ Probability gatecse-2002 probability easy binomial-distribution + – Kathleen asked Sep 15, 2014 Kathleen 10.4k views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply Ayush Upadhyaya commented Apr 18, 2017 i reshown by Ayush Upadhyaya Nov 27, 2017 reply Follow Share Total outcomes - 24 (Because 4 coins are tossed simultaneously and each coin has 2 outcomes-either head or tail) Now out of this 16 outcomes, one will be all HHHH(all heads) and other will be all TTTT(all tails) rest 14 outcomes will have atleast one head and one tail. So, probability, favourable events/total outcome 14/16 = 7/8 18 votes 18 votes Kiyoshi commented Jan 5, 2022 reply Follow Share Probability of at least one head and one tail = 1- Probability of No (head and tail) = 1 – 2 / (2*2*2*2) {Numerator HHHH and TTTT and denominator all choices} = 7/8 1 votes 1 votes yuyutsu commented Jan 19 reply Follow Share P(atleast one head)=P(x)=15/16 P(atleast one tail)=P(y)=15/16 P(atleast one head or atleast one tail) = 1. Hence, by inclusion exclusion principle 1=15/16 + 15/16 - P(x.y) P(x.y) = 7/8 0 votes 0 votes Please log in or register to add a comment.
–1 votes –1 votes probability of getting head=p=1/2; by the formula p(x)=nCx*P^xq^(n-x) p(atleast one tail)=p(x>=1)==1-p(x<1)=4C0*(1/2)^4=1-1/16=15/16 p(atleast one head)=p(x>=1)==1-p(x<1)=4C0*(1/2)^4=1-1/16=15/16 so P(head&tail)=15/16*15/16=7/8 PRANAV M answered Jun 17, 2017 PRANAV M comment Share Follow See all 2 Comments See all 2 2 Comments reply sai007 commented Jun 24, 2017 reply Follow Share number of probabilities of 4 coins = 2⁴ = 16 We have a formula that P(x≥1)+P(y≥1) = 1-(P(x<1)+P(y<1)) = 1-(P(x=0)+P(y=0)) = 1-((1/16)+(1/16)) = 1-(2/16) = 7/8 2 votes 2 votes sidlewis commented Nov 20, 2018 reply Follow Share Simplest Approach! 2 votes 2 votes Please log in or register to add a comment.