If A Coin Is Tossed 5 Times, Using Binomial Distribution
If a coin is tossed 5 times, using binomial distribution find the probability of:
(a) Exactly
2 heads
(b)
At least 4 heads
SOLUTION:
(a) According to the problem:
Number of trials: n = 5
The probability
of head: p = 1/2 and the probability of tail, q =1/2
For exactly
2 heads ( x = 2 ) = 5C2 . (1/2)2 . (1/2)5-2
= 5! /
2!3! x ¼ x 1/8
= 10 x
1/32
= 5/16
Answer; P
( x = 2 ) = 5/16.
(b)
Number of trials; n = 5
The probability
of head: p = 1/2 and the probability of tail, q =1/2
At least
4 heads (x > 4) = P ( x = 4 ) + P ( x = 5 )
P ( x = 4
) = 5C4 . (1/2)4 . (1/2)5-4
P ( x = 4
) = 5! / 4!1! x 1/16 x ½
P ( x = 4
) = 5/32
P ( x = 5
) = 5C5 . (1/2)5
. (1/2)5-5
P ( x = 5
) = 5! / 5!0! x 1/32 x 1
P ( x = 5
) = 1/32
P ( x >
4 ) = 5/32 + 1/32 = 6/32 = 3/16
Answer; P
( x > 4 ) = 3/16
Post a Comment for "If A Coin Is Tossed 5 Times, Using Binomial Distribution"