Question

Ten coins are thrown simultaneously. find the probability of getting at least 7 heads

Answer 1

P(getting a head) = p = 1/2
q = 1-p = 1-1/2 = 1/2
n = 10
The general form of binomial distribution for determining the probability of r successes is
P(r) = nCr*q^(n-r)*p^r
Required probability = P(r >/= 7) = P(r=7)+P(r=8)+P(r=9)+P(r=10)
Since p=q=1/2,
q^(n-r)*p^r = (1/2)^10
P(r >/= 7) = (1/2)^10*(10C7+10C8+10C9+10C10)
= 1/1024*(120+45+10+1)
= 1/1024*(176)
= 176 / 1024
= 0.171875

Answer 2

Given,

Ten coins are tossed simultaneously

n = 10

p = 1/2

q= 1/2

To find,

Probability of getting at least 7

Solution,

The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.

In the given case,

The probability of getting at least 7 when ten coins are tossed simultaneously is such that:

P(X>=7)=P(X=7)+P(X=8)+P(X=9)+P(X=10)

=(10C3+10C2+10C1+1)(1/2^10)

=176/(2^10)

=0.171875

Hence, when ten coins are tossed simultaneously the probability of getting at least 7 is 0.171875.