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3 years ago

If Miranda flips a coin twice, what is the probability that both coins will land heads up?

2 answers:

8 0

The probability that both coins will land heads up is 1/4. The probabilty the first coin will land heads up is 1/2. The probability that the second coin will land heads up is 1/2.

P(both land heads up) = (1/2)(1/2) = 1/4.

Vladimir79 [104]3 years ago

5 0

Answer:

Both heads: ¼

Step-by-step explanation:

First lands heads up: ½

Second lands heads up: ½

Both heads: ½ × ½ = ¼

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Please help me solve this

Viktor [21]

You can factor the 32 out of the sum:

$\displaystyle \sum_{m=1}^\infty 32 \cdot \left(\dfrac{1}{2}\right)^{m-1} = 32\sum_{m=1}^\infty \left(\dfrac{1}{2}\right)^{m-1}$

We can also change the index as follows

$\displaystyle 32\sum_{m=1}^\infty \left(\dfrac{1}{2}\right)^{m-1} = 32\sum_{m=0}^\infty \left(\dfrac{1}{2}\right)^{m}$

Now, we have a theorem that states that the series

$\displaystyle \sum_{m=1}^\infty a^m$

converges if and only if $|a|$ , and in this case we have

$\displaystyle \sum_{m=1}^\infty a^m = \dfrac{1}{1-a}$

This is your case, because you have

$|a|=\dfrac{1}{2}$

which implies that your series converges, and the value is

$\displaystyle 32\sum_{m=0}^\infty \left(\dfrac{1}{2}\right)^{m} = 32 \cdot \dfrac{1}{1-\frac{1}{2}} = 32\cdot 2 = 64$

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