# 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 ) = _{5}C_{2} . (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
) = _{5}C_{4} . (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
) = _{5}C_{5} . (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**

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