+7 votes
1.4k views
Four fair coins are tossed simultaneously. The probability that at least one head and one tail turn up is

A. $\frac{1}{16}$

B. $\frac{1}{8}$

C. $\frac{7}{8}$

D. $\frac{15}{16}$
asked | 1.4k views

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

## 4 Answers

+15 votes
Best answer

Answer - C

probability of getting all heads =$\dfrac{1}{16}$

probability of getting all tails =$\dfrac{1}{16}$

probability of getting at least one head and one tail $= 1 - \dfrac{1}{16} - \dfrac{1}{16} = \dfrac{7}{8}.$

answered by Boss (9.3k points)
edited
+10 votes

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

answered by Boss (7.6k points)
–1 vote
1 head 3 tails =1/16 2 heads 2 tails =1/16 3 heads 1 tail =1/16 adding we get 3/16 where i am wrong anyone plz explain
answered by Veteran (14.3k points)

1 head 3 tails =1/16 * 4C1 = 1/4

2 heads 2 tails =1/16 *4C2 = 3/8

3 heads 1 tail =1/16 *4C3 = 1/4

So, adding gives 7/8.

–1 vote
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
answered by (23 points)
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
Answer:

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